Strictly as per GTU syllabus Linear Algebra and Vector...

Linear Algebraand

Vector CalculusLAVC (GTU Subject Code - 2110015)

B.E. Semester IIVersion 1.0

(January 2018)

Powered by

B. Sc. (Gold Medalist)

M. Sc. (Gold Medalist)

Tetra Gold Medalist in Mathematics

Ph. D. (Mathematics)

ISTE (Reg. 107658)

IMS, AMS

Prof. (Dr.) Rajesh M. Darji

Target AAREAD | REDO | RECALL

Strictly as per GTU syllabus...

Latest version available at http://rmdarji.ijaamm.com/

Department of Mathematics

Sarvajanik College of Engg. & Tech. (SCET) SURAT

Dear Readers,

For any query regarding this subject

feel free to ask or WhatsApp on (+91) 9427 80 9779

Linear Algebraand

Vector CalculusLAVC (GTU Subject Code - 2110015)

B.E. Semester IIVersion 1.0

(January 2018)

Powered by

B. Sc. (Gold Medalist)

M. Sc. (Gold Medalist)

Tetra Gold Medalist in Mathematics

Ph. D. (Mathematics)

ISTE (Reg. 107658)

IMS, AMS

Prof. (Dr.) Rajesh M. Darji


Strictly as per GTU syllabus...

Latest version available at http://rmdarji.ijaamm.com/





Dear Readers,

For any query regarding this subject

feel free to ask or WhatsApp on (+91) 9427 80 9779


Powered by Prof. (Dr.) Rajesh M. Darji

Linear Algebraand

Vector CalculusLAVC (GTU- 2110015) B.E. Semester II

Dedicated toMy Beloved Students



Contents

1 Review of Matrices 11.1 Matrix . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 11.2 Types of Matrices . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1

1.2.1 Row and Column Matrix . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 11.2.2 Zero or Null Matrix . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 21.2.3 Square Matrix . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 21.2.4 Transpose of Matrix . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 21.2.5 Symmetric Matrix . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 21.2.6 Skew Symmetric Matrix . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 21.2.7 Diagonal Matrix . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 21.2.8 Scalar Matrix . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 31.2.9 Unit Matrix . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 31.2.10 Upper Triangle Matrix . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 31.2.11 Lower Triangle Matrix . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 3

1.3 Determinant of Matrix . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 31.4 Minor of an Element . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 31.5 Cofactor of an Element . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 31.6 Adjoint of a Matrix . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 41.7 Singular and Non-singular Matrix . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 41.8 Operations of Matrices . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 4

1.8.1 Equality . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 41.8.2 Multiplication of Matrix by a Scalar . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 41.8.3 Addition and Substation . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 41.8.4 Multiplication of Matrices . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 4

1.9 Inverse of a Matrix . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 51.10 Elementary Transformations on Matrix . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 51.11 Equivalent Matrices . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 51.12 Gauss-Jordan Method to find Inverse Matrix . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 61.13 Rank of a Matrix . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 91.14 Rank by Row Echelon Method: (Elementary Transformation Method) . . . . . . . . . . . . . . . 9

1.14.1 Row-Echelon or Canonical form . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 91.15 Reduced Row Echelon Form . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 9

2 System of Linear Algebraic Equations 122.1 System of Equations . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 122.2 Augmented Matrix . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 132.3 Non-Homogeneous System of Equations . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 132.4 Homogeneous System of Equations . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 132.5 Conditions for the Consistency of Non-Homogeneous System of Equations . . . . . . . . . . . 132.6 Conditions for the Consistency of the System of Homogeneous Equations . . . . . . . . . . . . 17

i



Prof. (Dr.) Rajesh M. Darji (Tetra Gold Medalist in Mathematics) ii

3 Notions of Vectors in Rn 203.1 Euclidean Space . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 203.2 Linear Combination . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 213.3 Linearly Independent Vectors (LI) . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 213.4 Linearly Dependent Vectors (LD) . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 213.5 Euclidean Norm . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 213.6 Normalized Vector . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 223.7 Euclidean Distance and Angle . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 223.8 Cauchy-Schwarz’s inequality . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 223.9 Minkowski’s Triangular Inequality . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 22

4 Vector Space 264.1 Field . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 264.2 Vector Space . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 264.3 Some Standard Vector Spaces . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 274.4 Subspace . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 304.5 Linear Combination and Span . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 324.6 Linearly Independent Vectors (LI) . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 324.7 Linearly Dependent Vectors (LD) . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 334.8 Wronskian . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 334.9 Basis . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 364.10 Some Standard Basis . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 374.11 Dimension . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 374.12 Ordered Basis and Coordinate Vector . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 414.13 Translation Matrix (Change of Basis Matrix) . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 424.14 Fundamental Spaces: Row Space, Column Space, Null Space . . . . . . . . . . . . . . . . . . . . 454.15 Rank and Nullity . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 464.16 Rank-Nullity Theorem . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 46

5 Linear Transformation (Linear Mapping) 515.1 Linear transformation . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 515.2 Particular Transformations . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 515.3 Matrix Linear Transformation . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 535.4 Composition Linear Transformations . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 545.5 Onto (Surjective) Linear Transformations . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 555.6 One-one (Injective) Linear Transformations . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 555.7 Range (Image) and Kernel . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 565.8 Inverse Linear Transformation (Isomorphism) . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 59

6 Eigenvalues and Eigenvectors 616.1 Definition . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 616.2 Method of Finding Eigenvalue and Eigenvector . . . . . . . . . . . . . . . . . . . . . . . . . . . . 616.3 Properties of Eigenvalues . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 616.4 Properties of Eigenvectors . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 626.5 Algebraic and Geometric Multiplicity of an eigenvalue . . . . . . . . . . . . . . . . . . . . . . . . 706.6 Cayley-Hamilton Theorem . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 726.7 Similar Matrices . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 746.8 Diagonalization . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 746.9 Orthogonally Diagonalization . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 74

LAVC (GTU-2110015) B.E. Semester II



Prof. (Dr.) Rajesh M. Darji (Tetra Gold Medalist in Mathematics) iii

7 Quadratic Forms and Complex Matrices 787.1 Quadratic Form (QF) . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 787.2 Matrix of Quadratic Form . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 787.3 Index, Signature and Rank of Quadratic Form . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 797.4 Definiteness of Quadratic Form . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 797.5 Complex Matrix . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 837.6 Conjugate Matrix . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 837.7 Conjugate Transpose . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 837.8 Hermitian, Skew-Hermitian, Unitary and Normal Matrices . . . . . . . . . . . . . . . . . . . . . 837.9 Properties . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 84

8 Inner Product Space and Orthogonal Basis 868.1 Inner Product Space . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 868.2 Properties of Inner Product . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 868.3 Some Standard Inner Product Spaces . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 868.4 Norm, Distance and Angle . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 888.5 Results . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 898.6 Orthogonal Complement . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 918.7 Properties of W⊥ . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 928.8 Results . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 928.9 Orthogonal Set . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 938.10 Orthogonal Projection . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 948.11 Orthogonal and Orthonormal Bases . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 958.12 Coordinate Relative to Orthonormal Basis . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 958.13 Gram-Schmidt Process . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 968.14 Least Square Approximate Solution for Linear System . . . . . . . . . . . . . . . . . . . . . . . . 1008.15 Orthogonal Matrix . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 101

9 Vector calculus I: Vector differentiation 1049.1 Scalar and Vector . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1049.2 Algebraic Operations of Vectors . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1049.3 Point Functions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1059.4 Vector Differential Operator . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1059.5 Gradient . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1059.6 Divergence . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1079.7 Curl . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1089.8 Directional Derivative . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1109.9 Angle between two Surfaces . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 111

10 Vector Calculus II: Vector Integration 11410.1 Line Integral . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 11410.2 Surface Integral (Normal Surface Integral) . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 11810.3 Volume Integral . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 12010.4 Integral Theorems . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 121




Chapter 1Review of Matrices

1.1 Matrix

A matrix is a rectangular arrangement of certain numbers (called elements or entries) in an array of m rowsand n columns such as,

A =

a 11 a 12 a 13 . . . . . . a 1n

a 21 a 22 a 23 . . . . . . a 2n

a 31 a 32 a 33 . . . . . . a 3n...

...

...

...

...

...

...

...

...

...

...

...a m1 a m2 a m3 . . . . . . a mn

is called an m ×n matrix and generally it is denoted by Am×n . Here m ×n is known as order of the matrix.More generally matrix can be denoted by, A m×n = [

a i j]

, Where i = 1,2,3, ...m, and j = 1,2,3, ...n.

Here a i j denotes the elements on the i th row and the j th column that may be real or complex.e. g.

A 3×2 = 1 2

2 63 −4

, A 3×3 = 3 0 4

−1 0.8 52 2 3+7i

Remark:

1. Distinct notations are used for enclosing the elements of matrix are [] , () , , ‖‖2. Elements a11, a22, a33... are said to be on leading diagonal or principal diagonal elements of the ma-

trix.

1.2 Types of Matrices

1.2.1 Row and Column Matrix

A matrix of order 1×n is having only one row and n column is known as row matrix or row vector.That is A 1×n = [

a11 a12 a13 . . . . a1n]

e. g.A 1×4 =

[ −2 1 0 3]

Similarly, a matrix of order m × 1 is having m rows and only one column is known as column matrix orcolumn vector.

That is A m×1 =

a11

a21

a31...

am1

e. g. A 3×1 = 0

12

1



Prof. (Dr.) Rajesh M. Darji (Tetra Gold Medalist in Mathematics) 2

1.2.2 Zero or Null Matrix

A matrix containing all zero elements is said to be zero matrix or null matrix and is denoted by Z or O.e. g.

Z 3×4 = 0 0 0 0

0 0 0 00 0 0 0

= O

1.2.3 Square Matrix

A matrix containing same numbers of rows and columns i.e. m = n is said to be square matrix. If A is squarematrix of order n then it is also denoted by An .e. g.

A 2×2 =[

0 1−1 2

]= A 2; A 3×3 =

1 2 30 −5 11 2 −2

= A 3

1.2.4 Transpose of Matrix

Matrix obtained by interchanging the rows and columns of the given matrix A is called transpose of A and isdenoted by the symbol A′ or AT .e. g.

A =

1 2 32 1 40 1 21 1 −1

⇒ AT = 1 2 0 1

2 1 1 13 4 2 −1

1.2.5 Symmetric Matrix

A square matrix A is said to be symmetric matrix if A = AT .e. g.

A = 1 2 −1

2 5 3−1 3 7

and A = a h g

h b fg f c

Thus, in a symmetric matrix ai j = a j i ∀i , j

1.2.6 Skew Symmetric Matrix

A square matrix A is said to be skew symmetric matrix if A =−AT .Thus, in a skew symmetric matrix ai j =−a j i ∀i j . Note that the diagonal elements of a skew symmetric

matrix are always zero because ai i =−ai i ⇒ ai i = 0e. g.

A = 0 1 −2

−1 0 32 −3 0

1.2.7 Diagonal Matrix

If in a square matrix, all non-diagonal elements are zeros then it is called diagonal matrix.e. g.

A = 1 0 0

0 2 00 0 3





1.2.8 Scalar Matrix

If a diagonal matrix has all diagonal elements equal i.e. a11 = a22 = a33.... then it is called scalar matrix.e. g.

A = 7 0 0

0 7 00 0 7

1.2.9 Unit Matrix

A diagonal matrix of order n in which all the diagonal elements are unity (one) is called unit matrix of ordern and is denoted by In . Unit matrix is also called an identity matrix.e. g.

I2 =[

1 00 1

]amd I3 =

1 0 00 1 00 0 1

1.2.10 Upper Triangle Matrix

It is a square matrix in which all the elements below the principle diagonal are zero.e. g.

A = 3 1 −2

0 7 40 0 1

1.2.11 Lower Triangle Matrix

It is a square matrix in which all the elements above the principle diagonal are zer.e. g.

A = 1 0 0

3 3 02 1 5

1.3 Determinant of Matrix

If A is a square matrix then determinant of A is denoted by | A | or det(A).e. g.

A = 2 3 1

1 2 3−1 1 0

⇒ | A | = det(A) =∣∣∣∣∣∣

2 3 11 2 3−1 1 0

∣∣∣∣∣∣=−12.

1.4 Minor of an Element

The minor of an element of | A | is determinant obtained by omitting the row and column in which the ele-ment present. In general the minor of an element ai j is denoted byMi j .

e. g. If |A | =∣∣∣∣∣∣

a1 b1 c1

a2 b2 c2

a3 b3 c3

∣∣∣∣∣∣ then, M11 =∣∣∣∣ b2 c2

b3 c3

∣∣∣∣ , M21 =∣∣∣∣ b1 c1

b3 c3

∣∣∣∣ , M22 =∣∣∣∣ a1 c1

a3 c3

∣∣∣∣ , M33 =∣∣∣∣ a1 b1

a3 b3

∣∣∣∣1.5 Cofactor of an Element

The cofactor of an element ai j of | A | is denoted by A i j and is defined as A i j = (−1)i+ j Mi j

e. g. If |A| =∣∣∣∣∣∣

a11 a12 a13

a21 a22 a23

a31 a32 a33

∣∣∣∣∣∣ then, A 11 = (−1)1+1M11 =∣∣∣∣ a22 a23

a32 a33

∣∣∣∣ , A 21 = (−1)2+1M21 =−∣∣∣∣ a12 a13

a32 a33

∣∣∣∣LAVC (GTU-2110015) B.E. Semester II




1.6 Adjoint of a Matrix

Adjoint of a square matrix A is the transpose of the matrix formed by the cofactor of the elements the given

matrix A and is denoted by adj(A). That is, if A = a11 a12 a13

a21 a22 a23

a31 a32 a33

then adj(A) = A11 A21 A31

A12 A22 A32

A13 A23 A33

e. g.

A = 1 2 3

0 −1 −34 5 −4

⇒ adj(A) = 19 23 −3

−12 −16 34 3 −1

(Verify !)

1.7 Singular and Non-singular Matrix

For a square matrix A if | A | = 0 then it is called singular and if | A | 6= 0 then it is called non-singular matrix.

1.8 Operations of Matrices

1.8.1 Equality

Two matrices A and B of the same order are said to be equal if all the elements of A and B in the correspond-ing position are equal.e. g.

A =[

1 23 4

], B =

[1 23 4

]⇒ A = B

1.8.2 Multiplication of Matrix by a Scalar

For any scalar k if , A = [a i j

]then k A = [

ka i j]

, 1 É i É m,1 É j É n.

e. g.

If A =[

13

2−1

32

]then, 2A =

[26

4−2

64

]and (−1) A =−A =

[ −1−3

−21

−3−2

]

1.8.3 Addition and Substation

Let A = [ai j

]and B = [

bi j]

, 1 É i É m,1 É j É n then A±B = [ai j ±bi j

].

e. g. Let A =[

1 21 2

], B =

[2 34 5

]then A+B =

[3 55 7

]and A−B =

[ −1 −1−3 −4

]

1.8.4 Multiplication of Matrices

Let A = [ai j

]and B = [

b j k]

matrices of orderm ×n and n × p respectively then product AB exist and it is

m ×p matrix which is define as AB = [ci k ] where ci k =n∑

j=1ai j b j k .

â In order to find product of two matrices A and B , take row from first matrix A and column from secondmatrix B .

â Find the product of respective entries of row and column, and then add them.

â It gives the entry on corresponding row and column of the product matrix (AB).

â For example, in following matrices A and B , if we consider first row (R1) from matrix A and secondcolumn (C2) from matrix B then corresponding entry of product matrix AB lies on first row and sec-ond column and is given by (1×1)+ (2×2)+ (3×4) = 17.

â Similarly, we can find all other entries in product matrix AB .





A2×3 =[

1 2 32 3 4

], B3×3 =

2 1 1−1 2 13 4 1

⇒ (AB)2×3 =[

9 17 613 24 9

]Remark:

1. In general AB 6= B A.

2. AB = O does not imply A = O or B = Oe. g.

A =[

1 21 2

],B =

[2 6−1 −3

]⇒ AB =

[0 00 0

]= O

3. If AB = AC does not imply B =C 4. A (BC ) = (AB)C [Associative Law]

5. A (B +C ) = AB + AC and (A+B)C = AC +BC [Distribution Law]

6. k (AB) = (k A)B = A (kB)

7. If A be a square matrix thenA2 = A A, A3 = A2 A. In general Am+n = Am An , A0 = I .and(

Am)n = Amn

8. If A be a square matrix of order n then AIn = In A = A

9.(

A′)′ = A ,(A+B)′ = A′+B ′ and (AB)′ = B ′A′

10. | AB | = | A | |B | 11. A adj(A) = adj(A) A = | A | I

1.9 Inverse of a Matrix

For a non-singular square matrix A if there exist another non-singular matrix B such that AB = B A = I thenmatrix A is called invertible and matrix B is called inverse of A. It is denoted by B = A−1 and is given by

A−1 = 1

|A| adj(A)

Thus a square matrix A is invertible if |A| 6= 0, That is, A is non-singular In this case A A−1 = A−1 A = I

Remark:

1. (AB)−1 = B−1 A−1 2.(

A′)−1 = (A−1)′, (

A−1)−1 = A, I−1n = In

1.10 Elementary Transformations on Matrix

1. The interchange of i th and j th rows is denoted by Ri j or Ri ↔ R j .

The interchange of i th and j th columns is denoted by Ci j or Ci ↔C j

2. The multiplication of each element of i th row by a nonzero scalar k is denoted by kR i .

The multiplication of each element of i th column by a nonzero scalar k is denoted by kC i .

3. Multiplication of each element of i th row by a nonzero scalar k and adding corresponding element tothe j th row is denoted by R i j (k) or

(R j → R j +kR i

)Multiplication of each element of i th row by a nonzero scalar k and adding corresponding element tothe j th column is denoted by C i j (k) or

(C j →C j +kC i

)1.11 Equivalent Matrices

Two matrices A and B are said to be equivalent if one can be obtained from other one by applying the finitenumbers of elementary operations and they are denoted by A ∼ B or A → B .





1.12 Gauss-Jordan Method to find Inverse Matrix

The method of finding the inverse of a given matrix by elementary row transformations is called Gauss-Jordan method and can be apply as follow:

[A : I] ⇒ [I : A−1]

Illustration 1.1 Prove that any square matrix can be expressed as sum of symmetric and skew symmet-

ric matrices. Hence express the matrix

4 2 −31 3 −6−5 0 7

as such a sum of symmetric and skew symmetric

matrices.

Solution: Let A be the square matrix of order n. Hence,

2A =(A+ AT )+ (

A− AT )∴ A =1

2

(A+ AT )+ 1

2

(A− AT )= P +Q

(Say

)(1.1)

Now, we will see that P and Q are symmetric and skew-symmetric matrices respectively.

P =1

2

(A+ AT )

∴ P T =[

1

2

(A+ AT )]T

= 1

2

(A+ AT )T [

∵ (k A)T = k AT ]=1

2

(AT + (

AT )T)= 1

2

(AT + A

)= P

∴ P T =P

Therefor, P is a symmetric matrix.Also,

Q =1

2

(A− AT )

∴ QT =[

1

2

(A− AT )]T

= 1

2

(A− AT )T

=1

2

(AT − (

AT )T)= 1

2

(AT − A

)=− 1

2

(A− AT )=−Q

∴ QT =−Q

Therefore Q is a skew-symmetric matrix.Hence from (1.1), it is proved that every square matrix can be expressed as sum of symmetric and skew

symmetric matrices.

Now let A = 4 2 −3

1 3 −6−5 0 7

= P +Q, where

P = 1

2

(A+ AT )= 1

2

4 2 −31 3 −6−5 0 7

+ 4 1 −5

2 3 0−3 −6 7

= 1

2

8 3 −83 6 −6−8 −6 14

∴ P = 4 3/2 −4

3/2 3 −3−4 −3 7





and

Q = 1

2

(A− AT )= 1

2

4 2 −31 3 −6−5 0 7

− 4 1 −5

2 3 0−3 −6 7

= 1

2

0 1 2−1 0 −6−2 6 0

∴ Q = 0 1/2 1

−1/2 0 −3−1 3 0

Hence, 4 2 −3

1 3 −6−5 0 7

= 4 3/2 −4

3/2 3 −3−4 −3 14

+ 0 1/2 1

−1/2 0 −3−1 3 0

Illustration 1.2 Find inverse of the matrix

0 1 21 2 33 1 1

by using Gauss-Jordan method.

Solution: Consider the matrix,

[A : I ] = 0 1 2 1 0 0

1 2 3 0 1 03 1 1 0 0 1

In order to find inverse of a given matrix, we transform A to I in above matrix by applying only row opera-tions successively, as follow:

[A : I ] = 0 1 2 1 0 0

1 2 3 0 1 03 1 1 0 0 1

→ R12

∼ 1 2 3 0 1 0

0 1 2 1 0 03 1 1 0 0 1

→ R13 (−3)

∼ 1 2 3 0 1 0

0 1 2 1 0 00 −5 −8 0 −3 1

→ R21 (−2) ,R23 (5)

∼ 1 0 −1 −2 1 0

0 1 2 1 0 00 0 2 5 −3 1

→ R31 (1/2) ,R32 (−1)

∼ 1 0 0 1/2 −1/2 1/2

0 1 0 −4 3 −10 0 2 5 −3 1

→(

1

2

)R3

∼ 1 0 0 1/2 −1/2 1/2

0 1 0 −4 3 −10 0 1 5/2 −3/2 1/2

= [I : A−1]

Hence,

A−1 = 1/2 −1/2 1/2

−4 3 −15/2 −3/2 1/2

Exercise 1.1

1. Express the matrix

1 2 4−2 5 3−1 6 3

as such a sum of symmetric and skew symmetric matrices.





2. Prove that inverse of a matrix is unique.

3. Find adjoint of the matrix A = 1 2 3

0 1 22 0 1

. Also verify that A adj(A) = adj(A) A = | A | I

4. Show that

[λ 10 λ

]n

=[λn nλn−1

0 λn

], where n is a positive integer.

5. Show that∣∣adj(A)

∣∣= | A |n−1 , where A be a square matrix of order n. Hence deduce that∣∣adj

(adjA

)∣∣=| A |(n−1)2

6. Find the inverse of the following matrices by using Gauss-Jordan method (using row operations):

a.

2 1 33 1 21 2 3

b.

1 3 31 4 31 3 4

c.

−1 −3 3 −11 1 −1 02 −5 2 −3−1 1 0 1

7. Find the inverse of the following matrices by using adjoint method if exist:

a.

1 1 11 2 31 4 9

b.

1 2 −2−1 3 00 −2 1

8. If A and B are symmetric matrices then prove that AB is symmetric, provided A and B commute.

9. If A is a non-singular matrix of order n, then show that adj(adj A

)= | A |n−2 A. Hence find adj(adj A

), if

A = 1

9

−1 −8 4−4 4 7−8 −1 −4

Answers

1.

1 0 3/20 5 9/2

3/2 9/2 3

+ 0 2 5/2

−2 0 −3/2−5/2 3/2 0

3. adj(A) = 1 −2 1

4 −5 −2−2 4 1

4. Hint: Use principle of mathematical induction.

6. a.

−1/6 1/2 −1/6−7/6 1/2 5/65/6 −1/2 −1/6

b.

7 −3 −3−1 1 0−1 0 1

c.

0 2 1 31 1 −1 −21 2 0 1−1 1 2 6

7. a.

3 −5/2 1/2−3 4 −11 −3/2 1/2

b.

3 2 61 1 22 2 5

9. A

E E E





1.13 Rank of a Matrix

The matrix is said to be of rank r if there is,

1. At least one minor of the order r which is not equal to zero and

2. Every minor of order (r +1) is equal to zero.

The rank of a matrix A is denoted by ρ (A) = rRemark:

1. The rank of a matrix A is the maximum order of its non vanishing minor.

2. If the matrix A has non-zero minor of order r then ρ (A) Ê r

3. If the matrix A has all the minors of order (r+1) as zeros ρ (A) É r

4. If A is an m ×n matrix then ρ (A) É min(m,n)

5. Elementary transformations do not alter the order and the rank of the matrix.

1.14 Rank by Row Echelon Method: (Elementary Transformation Method)

1.14.1 Row-Echelon or Canonical form

â Let A be the matrix of order m ×n i. e. A m×n .

â The row-echelon or canonical form of the matrix A m×n is a matrix in which one or more elements ineach of the first r rows are non-zero and all the elements in remaining rows are zeroes.

â Any matrix A can always be reduced to the echelon form by applying only row transformations.

â In this case rank of a matrix is given by ρ (A) = m −k, where m denotes total numbers of rows and kdenotes total numbers of zero-rows. Note that if k = 0 then ρ (A) = m.

* Important:

â Pivot: In echelon form of matrix the first non-zero element of non-zero row from left is called Pivot ofthe corresponding row and the corresponding row is known as pivot row, as show in below matrix.

1 3 5 40 6 −1 40 0 0 20 0 0 0

â Observe that R1,R2,R3 are pivot rows whose pivots are enclosed by the rectangular box.

â Further the column on which the pivot of row exist is known as pivot column. In above matrixC1,C2,C4 are pivot columns.

1.15 Reduced Row Echelon Form

Reduced-echelon form is a row echelon form in which all the pivot are unity and above all elements of pivotare zeros.e. g.

1 0 −3 0 20 1 2 0 80 0 0 1 50 0 0 0 0





Illustration 1.3 Find the rank of the matrix

1 2 31 4 22 6 5

, using determinant method.

Solution: Let A = 1 2 3

1 4 22 6 5

Obviously, the highest ordered minor of A is 3rd order and it is det(A) itself.

∴ det(A) =∣∣∣∣∣∣

1 2 31 4 22 6 5

∣∣∣∣∣∣= 1(20−12)−2(5−4)+3(6−8) = 8−2+6 = 0.

So rank of A is not 3.

Now, consider 2nd order minor of A formed by its 1st and 2nd rows as

∣∣∣∣ 1 21 4

∣∣∣∣= 4−2 = 2 6= 0.

Hence, rank of matrix A is 2. ∴ ρ(A) = 2.

Illustration 1.4 Find the rank of the matrix

0 1 −3 −11 0 1 13 1 0 21 1 −2 0

by reducing to echelon echelon form.

Solution: Let A =

0 1 −3 −11 0 1 13 1 0 21 1 −2 0

To reduce A to its row-echelon form, use only row transformations, and bring 0 (zero) under the first nonzero (pivot) element of each row (enclosed by rectangular box in bellow process), starting from first row.

A =

0 1 −3 −11 0 1 13 1 0 21 1 −2 0

R3 → R3 −R1;R4 → R4 −R1

∼

0 1 −3 −11 0 1 13 0 3 31 0 1 1

R3 → R3 −3R2;R4 → R4 −R2

∼

0 1 −3 −11 0 1 10 0 0 00 0 0 0

Hence, ρ(A) = m −k = 4−2 = 2. (number of non-zero rows)

Exercise 1.2

1. Find the rank of the matrix

1 2 −1 33 4 0 −1−1 0 −2 7

using determinant method.

2. Find the rank of the following matrices by reducing to the echelon form:

a.

2 3 −1 −11 −1 −2 −43 1 3 −26 3 0 −7

b.

1 2 −2 32 5 −4 6−1 −3 2 −22 4 −1 6

3. Convert the matrix

1 2 32 3 43 4 5

in to reduced row echelon form and hence find the rank of matrix.





Answers

1. 2 2. a. 3 b. 4 3.

1 0 −10 1 120 0 0

, Rank = 2

E E E

Powered by

Prof. (Dr.) Rajesh M. DarjiB. Sc. (Gold Medalist)

M. Sc. (Gold Medalist)Tetra Gold Medalist in Mathematics

Ph. D. (Mathematics)ISTE (Reg. 107058)

IMS, AMShttp://rmdarji.ijaamm.com/

Contact: (+91) 9427 80 9779


http://rmdarji.ijaamm.com/



Chapter 2System of Linear Algebraic Equations

2.1 System of Equations

The collection of more than one linear equation is called system of equations.Consider the system of m equations in n unknown x 1, x 2, x 3.....x n as follow:

a 11x 1 +a 12x 2 +a 13x 3 + . . . . . .+a 1n x n = b 1

a 21x 1 +a 22x 2 +a 23x 3 + . . . . . .+a 2n x n = b 2

a 31x 1 +a 32x 2 +a 33x 3 + . . . . . .+a 3n x n = b 3

· · · · · · · · · · · · · · · · · · · · · · · · · · ·· · · · · · · · · · · · · · · · · · · · · · · · · · ·a m1x 1 +a m2x 2 +a m3x 3 + . . . . . .+a mn x n = b m

The solutions of above system means the value of unknown x 1, x 2, x 3.....x n that will satisfy the given system,that may or may not be exist.â The above system can be rewrite in compact form by using matrix notations as,

a 11 a 12 a 13 . . . . . . a 1n

a 21 a 22 a 23 . . . . . . a 2n

a 31 a 32 a 33 . . . . . . a 3n...

...... . . . . . .

......

...... . . . . . .

...a m1 a m2 a m3 . . . . . . a mn

x 1

x 2

x 3......

x n

=

b 1

b 2

b 3......

b m

That is, AX = B , where

A =

a 11 a 12 a 13 . . . . . . a 1n

a 21 a 22 a 23 . . . . . . a 2n

a 31 a 32 a 33 . . . . . . a 3n...

... . . . . . ....

...... . . . . . .

...a m1 a m2 a m3 . . . . . . a mn

; X =

x 1

x 2

x 3......

x n

; B =

b 1

b 2

b 3......

b n

* Important:

When the system AX = B has the solution then system is said to be consistent otherwise the system is calledinconsistent.

12




2.2 Augmented Matrix

If AX = B is the system of m equations in n unknown then the matrix written as [A : B ] or (A,B) is called asthe Augmented matrix of the given system. Hence

[A : B ] =

a 11 a 12 a 13 . . . . . . a 1n b 1

a 21 a 22 a 23 . . . . . . a 2n b 2

a 31 a 32 a 33 . . . . . . a 3n b 3...

...... . . . . . .

......

......

... . . . . . ....

...a m1 a m2 a m3 . . . . . . a mn b m

2.3 Non-Homogeneous System of Equations

For the system of equation AX = B , if matrix B is not a null matrix (non-zero matrix) then the system ofequation AX = B is known as non-homogeneous system of equations.

2.4 Homogeneous System of Equations

For the system of equation AX = B , if matrix B is a null matrix (zero matrix) then the system of equationAX = B (AX = Z ) is known as Homogeneous system of equations.e. g.

x + y + z = 3

x − y +2z = 4

2x +3y − z = 0

Non - homo. equations;

x + y + z = 0

2x +3y −4z = 0

x − y +2z = 0

Homo. equations.

2.5 Conditions for the Consistency of Non-Homogeneous System of Equations

Consider the system non-homogeneous of m equations in n unknown as AX = B .For the augmented matrix [A : B ], if

1. ρ (A) 6= ρ (A : B), then the system is inconsistent and possesses no solution.

2. ρ (A) = ρ (A : B), then the system is consistent and possesses solution.

In this case,

â If ρ (A) = ρ (A : B) = n (number of unknown) then solution is unique, and

â If ρ (A) = ρ (A : B) = r < n, then system possess infinite numbers of solutions, and that can berepresented parametrically in terms of (n − r ) parameters.

3. In particular let m = n = 3, that is three equations in three unknown, then

â If | A | 6= 0 then the system has unique solution and is given by, AX = B ⇒ X = A−1B .

â If | A | = 0 then the system is inconsistent or has infinite numbers of solutions, which can befollowed by reduction method.

* Important:

â To find rank of the augmented matrix we shall reduce the matrix in to the row echelon form.

â This method is known as Reduction Method or Gauss Elimination Method. Also in this method weapply only row elementary transformations.

â Gauss Jordan Elimination Method: In this method convert the matrix system [A : B ] in to reducedrow echelon form and then apply back substitution.





Illustration 2.1 Examine the consistency of following system of equations, and solve if consistent by Gausselimination method:

x1 +x2 +x3 = 9, 2x1 +4x2 −3x3 = 1, 3x1 +6x2 −5x3 = 0.

Solution: Here given system has three equations in three unknown xA , x2, and x3, that is (m ×n = 3×3)system.

In Gauss elimination method, we reduce augmented matrix [A : B ] to row echelon form using row oper-ations only, as follow:

[A : B ] = 1 1 2 9

2 4 −3 13 6 −5 0

R2 → R2 −2R1;R3 → R3 −3R1

∼ 1 1 2 9

0 2 −7 −170 3 −11 −27

R3 → R3 − 3

2R2

∼ 1 1 2 9

0 2 −7 −170 0 −1/2 −3/2

(2.1)

Observe that, ρ(A) = ρ(A : B) = 3 = n (number of unknown x1, x2, x3).∴ Given system of equation is consistent and has unique solution, and is given by making back substitu-tion from (2.1), as

R3 : −1

2x3 =−3

2R2 : 2x2 −7x3 =−17

R1 : x1 +x2 +2x3 = 9

⇒ x1 = 1, x2 = 2, x3 = 3.

Illustration 2.2 Test the consistency of the following equations and solve them if they consistent, by Gausselimination method:

2x +2y +2z = 0, −2x +5y +2z = 1, 8x + y +4z =−1.

Solution: The augmented matrix for given (m ×n = 3×3) system is,

[A : B ] = 2 2 2 0

−2 5 2 18 1 4 −1

R2 → R2 +R1;R3 → R3 −4R1

∼ 2 2 2 0

0 7 4 10 −7 −4 −1

R3 → R3 +R2

∼

2 2 2 00 7 4 10 0 0 0

(2.2)

Observe that, ρ(A) = ρ(A : B) = 2 < (n = 3).∴ Given system is consistent and has infinite solutions, which can be expressed in terms of one parameter(n − r = 3− 2 = 1). Such solutions are known as 1-parametric solution and can be obtained by assumingarbitrary parameter say t ∈R, to unknown corresponding to non-pivot column. Such unknown is called freevariable. In (2.2), z is a free variable. Hence, making back substitution from (2.2), we get

z = t , t ∈RR2 : 7y +4z = 1 ∴ y = 1−4t

7,

R1 : 2x +2y +2z = 0 ∴ x =−1+3t

7





Thus required solution is

x =−1+3t

7, y = 1−4t

7, z = t , t ∈R

Illustration 2.3 Solve the following system of equations by Gauss-Jordan elimination:

x1 +3x2 −2x3 +2x5 = 0

2x1 +6x2 −5x3 −2x4 +4x5 −3x6 = 1

5x3 +10x4 +15x6 = 5

2x1 +6x2 +8x4 +4x5 +18x6 = 6

Solution: Given system is of four equatons in six unknowns x1, x2, x3, x4, x5, x6. That is (4×6).In Gauss-Jordan elimination method, we reduce the augmented matrix to reduced row echelon form

using only row operations, as

[A : B ] =

1 3 −2 0 2 0 02 6 −5 −2 4 −3 −10 0 5 10 0 15 52 6 0 8 4 18 6

R2 → R2 −2R1;R4 → R4 −2R1

∼

1 3 −2 0 2 0 00 0 −1 −2 0 −3 −10 0 5 10 0 15 50 0 4 8 0 18 6

R3 → R3 +5R2;R4 → R4 +4R2

∼

1 3 −2 0 2 0 00 0 −1 −2 0 −3 −10 0 0 0 0 0 00 0 0 0 0 6 2

R2 → (−1)R2;R3 ↔ R4 (i.e R34)

∼

1 3 −2 0 2 0 00 0 1 2 0 3 10 0 0 0 0 6 20 0 0 0 0 0 0

R4 →(

1

6

)R4

∼

1 3 −2 0 2 0 00 0 1 2 0 3 10 0 0 0 0 1 1/30 0 0 0 0 0 0

R1 → R1 +2R2;R2 → R2 −3R3

∼

1 3 0 2 2 3 20 0 1 2 0 0 00 0 0 0 0 1 1/30 0 0 0 0 0 0

(Reduced row echelon form) (2.3)

∴ ρ (A) = ρ (A,B) = 3 < 6 (= Number of unknowns)

Therefore, system has (6−3 = 3) 3-parametric infinite solutions, and three parameters say r, s, t are assignto free variables x2, x4, x5 respectively, that is x2 = r, x4 = s, x5 = t . Thus from (2.3), required solution is givenby

x1 =−3r −4s −2t , x2 = r, x3 =−2s, x4 = s, x5 = t , x6 = 1

3, r, s, t ∈R

Illustration 2.4 For what values of λ and µ do the system of equations

x + y + z = 6, x +2y +3z = 10, x +2y +λz =µ,

has (i) a unique solution, (ii) an infinite numbers of solutions and (iii) no solution.





Solution: Consider the augmented matrix [A : B ] for given equations and apply reduction method, as

[A : B ] = 1 1 1 6

1 2 3 101 2 λ µ

R2 → R2 −R1;R3 → R3 −R1

∼ 1 1 1 6

0 1 2 40 1 λ−1 µ−6

R3 → R3 −R2

∼ 1 1 1 6

0 1 2 40 0 λ−3 µ−10

Observe that,

i. If λ 6= 3 then ρ(A) = ρ(A : B) = 3 ∀µ ∈R. Hence system has unique solution.

ii. If λ= 3 and µ= 10, than ρ(A) = ρ(A : B) = 2 < 3. Hence system has 1-parametric infinite solutions.

iii. If λ = 3 and µ 6= 10, than ρ(A) = 2,ρ(A : B) = 3, that is ρ(A) 6= ρ(A : B). Hence system is inconsistentand has no solution.

Exercise 2.1

1. Examine the following system of equations for consistency and if consistent then solve it, using Gausselimination method:

a. 2x1 +x2 −x3 +3x4 = 8, x1 +x2 +x3 −x4 =−2, 3x1 +2x2 −x3 = 6, 4x2 +3x3 +2x4 =−8.

b. x + y + z = 6, x − y +2z = 3, 3x + y + z = 8, 2x −2y +3z = 7.

c. 4x −2y +6z = 8, x + y −3z =−1, 15x −3y +9z = 21.

2. Solve the following equations using Gauss Jordan elimination method, if they consistent:

2x1 +x2 −x3 +3x4 = 11, x1 −2x2 +x3 +x4 = 8, 4x1 +7x2 +2x3 −x4 = 0, 3x1 +5x2 +4x3 +4x4 = 17.

3. Solve the following system of equations for x, y, z:

−1

x+ 3

y+ 4

z= 30,

3

x+ 2

y− 1

z= 9,

2

x− 1

y+ 2

z= 10.

[Hint: Put1

x= u,

1

y= v,

1

z= w.]

4. Find for what value of λ the set of equations,

2x −3y +6z −5w = 3, y −6z +w = 1, 4x −5y +6z −9w =λ,

has (i) no solution, (ii) infinite numbers of the solutions.

5. Show that if µ 6= 0 then the system of equations,

2x + y = a, x +µy − z = b, y +2z = c

has unique solution for all a,b,c. Also if µ= 0 then determine the relation satisfied by a,b,c such thatsystem is inconsistent. Find the general solution by taking µ= 0, a = 1,b = 1,c =−1.

6. Investigate for what values of a and b the system of simultaneous equations:

2x − y +3z = 6, x + y +2z = 2, 5x − y +az = b,

has (i) no solution, (ii) a unique solution and (iii) an infinite solutions.





7. Solve the following system by matrix inversion method:

x + y + z = 0, 2x +3y − z =−5, x − y + z = 4.

[Hint: AX = B ⇒ X = A−1B ]

8. Use matrix inversion method to determine the value of λ for which following system is consistent:

x +2y + z = 3, x + y + z =λ, 3x + y +3z =λ2,

Answers

1. a. x1 = 2, x2 =−1, x3 =−2, x4 = 1 b. Inconsistent c. x = 1, y = 3t −2, z = t , t ∈R

2. x1 = 2, x2 =−1, x3 = 1, x4 = 3 3. x = 1

2, y = 1

4, z = 1

54. (i) λ 6= 7 (ii) λ= 7

5. a = 2b + c, x = 1+ t , y =−1−2t , z = t , t ∈R 6. (i) a = 8,b = 6 (ii) a 6= 8,b ∈R (iii) a 6= 8,b 6= 6

7. x = 1, y =−2, z = 1 8. λ= 2,3

E E E

2.6 Conditions for the Consistency of the System of Homogeneous Equations

Consider the homogeneous system of m equations in n unknown as AX = Z , where Z is a null matrix oforder (m ×1).

For the augmented matrix [A : Z ], we note that ρ (A) = ρ (A : Z ). Hence Homogeneous system is alwaysconsistent and has either unique solution or infinite numbers solutions. This can be written as follow:

1. If ρ (A) = ρ (A : Z ) = n, (number of unknown) then the system possess unique solution and is given byx1 = x2 = x3 = .... = xn = 0, which is also known as trivial solution.

2. If ρ (A) = ρ (A : Z ) = r < n, then system possess infinite numbers of solutions which can be repre-sented parametrically in trems of (n-r)-parameters. These solutions are are also known as non-trivialsolutions.

3. In particular let m = n = 3, that is three equations in three unknown, then

â If |A| 6= 0, the system has unique solution (trivial) and is given by x1 = x2 = x3 = 0.

â If |A| = 0, the system has infinite numbers of parametric solutions (non-trivial) which can befollowed by reduction method.

Illustration 2.5 Solve the equations:

x1 +3x2 +2x3 = 0, 2x1 −x2 +3x3 = 0, 3x1 −5x2 +4x3 = 0, , x1 +17x2 +4x3 = 0.

Solution: The augmented matrix is

[A : Z ] =

1 3 2 02 −1 3 03 −5 4 01 17 4 0

→ R2 −2R1; R3 −3R1; R4 −R1

∼

1 3 2 00 −7 1 00 −14 −2 00 14 2 0

→ R3 −2R2; R4 +2R1





∼

1 3 2 00 −7 −1 00 0 0 00 0 0 0

∴ ρ (A) = ρ (A : Z ) =2 < 3 (number of unknowns)

Hence system has an infinite numbers of non-trivial (1-parametric) solutions, given by

x1 = 11t

7, x2 =− t

7, x3 = t , t ∈R

Illustration 2.6 Find the values of k for which the system of equations (3k−8)x+3y+3z = 0,3x+(3k−)y+3z = 0,3x +3y + (3k −8)z = 0 has a non-trivial solution.

Solution: For the given system of equations to have a n0n-trivial solution, the determinant of the coeffi-cient matrix should be zero. That is,∣∣∣∣∣∣

3k −8 3 33 3k −8 33 3 3k −8

∣∣∣∣∣∣= 0

⇒∣∣∣∣∣∣

3k −2 3k −2 3k −23 3k −8 33 3 3k −8

∣∣∣∣∣∣= 0[Operating R1 +R2 +R3

]

⇒ (3k −2)

∣∣∣∣∣∣1 1 13 3k −8 33 3 3k −8

∣∣∣∣∣∣= 0

⇒ (3k −2)

∣∣∣∣∣∣1 0 03 3k −11 33 3 3k −11

∣∣∣∣∣∣= 0[Operating C2 −C1; C3 −C1

]⇒ (3k −2)(3k −11)2 = 0

⇒ k = 2

3,

11

3,

11

3

Exercise 2.2

1. Solve the following system of equations:

a. 2x1+x2+3x3+6x4 = 0, 3x1−x2+x3+3x4 = 0, −x1−2x2+3x3 = 0, −x1−4x2−3x3−3x4 = 0.

b. x + y +2z = 0, x +2y +3z = 0, x +3y +4z = 0, x +4y +7z = 0.

c. x +2y +3z = 0, 2x +3y + z = 0, 4x +5y +4z = 0.

2. Examine the following system for the non-trivial solution:

5x +2y −3z = 0, 3x + y + z = 0, 2x + y +6z = 0.

3. For the different values of k discuss the nature of the solutions of the following system:

x +2y − z = 0, 3x + (k +7) y −3z = 0, 2x +4y + (k −3) z = 0.

4. Show that the system of equations, ax+by+cz = 0, bx+c y+az = 0, cx+ay+bz = 0, has a non-trivialsolution if a +b + c = 0 or a = b = c.





5. Show that the system of equations, x +2y +3z = λx, 3x + y +2z = λy, 2x +3y + z = λz, can possess anon-trivial solution only if λ= 6. Obtain the non-trivial solution for real value of λ.

6. Find the value of λ for which the equations,

(λ−1) x + (3λ+1) y +2λz = 0, (λ−1) x + (4λ−2) y + (λ+3) z = 0, 2x + (2λ+1) y +3(λ−1) z = 0,

are consistent, and find the ratio x : y : z when λ has smallest of these values. What happens when λ

has the greatest of these values.

Answers

1. a. x1 = x2 = x3 = x4 = 0 b. x = y = t , z =−t , t ∈R c. det(A) 6= 0, x = y = z = 0

2. system has non-trivial solution. 3. For k = 1, x = y = z = 0 ; For k 6= 1, x =−2t , y = t , z = 0, t ∈R5. x = y = z = t , t ∈R 6. λ= 0,3

E E E

Powered by





Contact: (+91) 9427 80 9779





Chapter 3Notions of Vectors in Rn

3.1 Euclidean Space 1

â Let R denotes the set of real numbers then, an n times Cartesian product of Rwith itself is denoted byRn . That is

Rn =R×R×R× ...×R (n times)

â In particular, R2 =R×R, R3 =R×R×Râ Elements of Rn are (x1, x2, ...xn) , xi ∈R ans are known as an order n − tuples. Thus

Rn = (x 1, x 2, x 3, . . . . . . x n) :x i ∈R, 1 É i É n

â Elements of R2 are called order pair and elements of R3 are called order triplets.

R2 = (x, y

): x, y ∈R

, R3 = (x, y, z

): x, y, z ∈R

â The elements of Rn are also referred as a vector or point and can be presented by mean of columnmatrix as

x =

x1

x2

x3...

xn

= [x1 x2 x3 . . . xn

]T = X

â Here Rn is known as Real Euclidean n dimensional space. For example, R2 and R3 are two and threedimensional spaces respectively.

â The elements x 1, x 2, x 3, ....x n are called components of the vector.

â The standard arithmetic addition, subtraction, scalar-multiplication, zero (null) vector etc. in Rn areas same as define for the matrices.

â The multiplication between two vectors in Rn define as follow:

1Euclid or Father of Geometry; Greek, Mid-4th century BCE-Mid-3rd century BCE.

20




Let x =

x 1

x 2

x 3...

x n

, y =

y1

y2

y 3...

y n

∈Rn then

yT x = [y1 y2 y3 · · · yn

]

x1

x2

x3...

xn

= x1 y1 +x2 y2 +x3 y3 + .....+xn yn

â This product is known as Euclidean Inner product or Dot product and is denoted by x · y . Hence,

x · y =n∑

i=1xi yi , 1 É i É n.

e. g. Let x = (1,4,−2) , y = (2,−1,3) ∈R3 ⇒ x · y = (1)(2)+ (4)(−1)+ (2)(3) = 2−4+6 = 4

3.2 Linear Combination

Let x 1, x 2, x 3, ...x n ∈Rn and c 1,c 2,c 3, ...c n ∈R then the vector,

c1x1 + c2x2 + c3x3 + .....+ cn xn =n∑

i=1ci xi

is called the linear combination of the vectors.e. g. Let x1 = (1,2,−1) , x2 = (2,1,1) , x3 = (1,0,3) ∈R3 then linear combination of x1, x2, x3 is

c1x1 + c2x2 + c3x3 =c1 (1,2,−1)+ c2 (2,1,1)+ c3 (1,0,3)

= (c1 +2c2 + c3,2c1 + c2,−c1 + c2 +2c3) , c1,c2,c3 ∈R

3.3 Linearly Independent Vectors (LI)

The vectors x1, x2, x3, ...xn ∈Rn are said to be linearly independent vectors if, whenever

c1x1 + c2x2 + c3x3 + .....+ cn xn = 0 ⇒ c1 = c2 = c3 = ..... = cn = 0.

3.4 Linearly Dependent Vectors (LD)

The vectors x1, x2, x3, ...xn ∈Rn are said to be linearly dependent vectors if, whenever

c1x1 + c2x2 + c3x3 + .....+ cn xn = 0 ⇒ Not all c1,c2,c3, .....cn are 0.

That is at least one constant is non-zero.â In this case at least one vector can always be expressed as a linear combination of rest of the vectors.

3.5 Euclidean Norm

Let x = (x 1, x 2, x 3, ...x n) ∈Rn then norm or magnitude of vector x is denoted by∥∥x

∥∥ and is defined as,∥∥x∥∥=

√x · x =

√x2

1 +x22 +x2

3 + .....+x2n

â The norm of a vector is also called length of a vector.

e. g. x = (1,2,−2,3) ∈R4 ⇒ ∥∥x∥∥=

√(1)2 + (2)2 + (−2)2 + (3)2 =p

1+4+4+9 =p18 = 3

p2





3.6 Normalized Vector

The vector of unit norm is called unit vector or normalized vector and is denoted byx.

â If∥∥x

∥∥ 6= 1 then x can be converted to normalized vector by dividing with∥∥x

∥∥. Thus, x = x∥∥x∥∥ be always

normalized vector.

e. g.[

cosθsinθ

],

[10

],

[0−1

]are normalized (unit) vectors in R2.

3.7 Euclidean Distance and Angle

Let x, y ∈Rn

â Distance: The distance between two vectors is defined as d(x, y

)= ∥∥x − y∥∥ .

â Angle: The angle θ between two vectors is defined as cosθ = x · y∥∥x∥∥∥∥ y

∥∥ .

Also, x, y ∈Rn are called orthogonal (perpendicular) vectors is cosθ = 0 i.e. x · y = 0

3.8 Cauchy-Schwarz’s inequality 2

For x, y ∈Rn ,∣∣x · y

∣∣É ∥∥x∥∥ ∥∥ y

∥∥Proof: Angle between two vectors x, y ∈Rn is defined as,

cosθ = x · y∥∥x∥∥∥∥ y

∥∥ ⇒ | cosθ | =∣∣∣∣ x · y∥∥x

∥∥∥∥ y∥∥

∣∣∣∣=∣∣x · y

∣∣∥∥x∥∥∥∥ y

∥∥But | cosθ | É 1,

∴

∣∣x · y∣∣∥∥x

∥∥∥∥ y∥∥ É 1,

Hence, ∣∣x · y∣∣É ∥∥x

∥∥∥∥ y∥∥

3.9 Minkowski’s Triangular Inequality 3

For x, y ∈Rn ,∥∥x + y

∥∥É ∥∥x∥∥+∥∥ y

∥∥ .

Proof: By definition of norm for x, y ∈Rn , we have∥∥x + y∥∥2 = (

x + y) · (x + y

)= (

x + y) · (x + y

)= x · x +x · y + y · x + y · y

= x · x +2x · y + y · y[∵ x · y = y · x

]É ∥∥x

∥∥2 +2∥∥x

∥∥∥∥ y∥∥+∥∥ y

∥∥2[∵

∣∣x · y∣∣É ∥∥x

∥∥∥∥ y∥∥]

∴∥∥x + y

∥∥2 É (∥∥x∥∥+∥∥ y

∥∥)2∥∥x + y∥∥É ∥∥x

∥∥+∥∥ y∥∥

2Augustin-Louis Cauchy; French, 1789 and Karl Hermann Amandus Schwarz; German, 1843-1921.3Hermann Minkowski; German, 1864-1909.





Illustration 3.1 Find the constant k such that the vectors (1,k,−3) and (2,−5,4) are orthogonal.

Solution: Let x = (1,k,−3) and y = (2,−5,4). For orthogonal vectors,

x · y = 0 ⇒ (1)(2)+ (k) (−5)+ (−3)(4) = 0

∴ 2−5k −12 = 0 ⇒ k =−2

Illustration 3.2 verify the Cauchy-Schwartz’s and triangle inequality for the vectors x = (1,−3,2) and y =(1,1,−1). Also find distance and angle between them.

Solution:

x = (1,−3,2) , y = (1,1−1) ⇒ x · y = 1−3−2 =−4,∥∥x

∥∥=p1+9+4 =p

14,∥∥y

∥∥=p1+1+1 =p

3

Therefore,∣∣x · y∣∣= |−5| = 5 and

∥∥x∥∥∥∥y

∥∥=p

14p

3 =p

42 ⇒ ∣∣x · y∣∣É ∥∥x

∥∥∥∥y∥∥ (3.1)

∥∥x + y∥∥= ‖(2,−2,1)‖ =

p4+4+1 = 3 and

∥∥x∥∥+∥∥y

∥∥=p

14+p3 ⇒ ∥∥x + y

∥∥É ∥∥x∥∥+∥∥y

∥∥ (3.2)

Hence from (3.1) and (3.2), Cauchy-Schwarz’s and Triangle angle inequalities are verified.Also distance and angle between them are given by

d(x, y

)= ∥∥x − y∥∥= ‖(0,−4,3)‖ =

p0+16+3

∴ d(x, y

)=p19

and

cosθ =x · y∥∥x∥∥∥∥y

∥∥ = (−5)p14

p3=− 5p

42

∴ θ = cos−1(− 5p

42

)

Illustration 3.3 Show that x1, x2, x3 are linearly independent and x4 depends on them, where x1 = (1,2,4) , x2 =(2,−1,3) , x3 = (0,1,2) , x4 = (−3,7,2).

Solution: For x1, x2, x3 consider the linear combination,

c1x1 + c2x2 + c3x3 = 0

∴ c1 (1,2,4)+ c2 (2,−1,3)+ c3 (0,1,2) = (0,0,0)

⇒ c1 +2c2 = 0, 2c1 − c2 + c3 = 0, 4c1 +3c2 +2c3 = 0. (3.3)

Now x1, x2, x3 are linearly independent if c1 = c2 = c3 = 0, that is the homogeneous system (3.3) should havetrivial solution. The determinant of coefficient matrix is∣∣∣∣∣∣

1 2 02 −1 14 3 2

∣∣∣∣∣∣= 1(−2−5)−2(4−4)+0 =−7 6= 0

∴ system (3.3) has trivial solution, hence x1, x2, x3 are linearly independent.Now including fourth vector x4 = (−3,4,7) in above linear combination, we get

c1x1 + c2x2 + c3x3 + c4x4 = 0

∴ c1 (1,2,4)+ c2 (2,−1,3)+ c3 (0,1,2)+ c4 (−3,7,4) = (0,0,0)





⇒ c1 +2c2 −3c4 = 0, 2c1 − c2 + c3 +7c4 = 0, 4c1 +3c2 +2c3 +4c4 = 0. (3.4)

Now we solve homogeneous system (3.4) for unknown c1,c2,c3 and c4. The augmented matrix is

[A : Z ] = 1 2 0 −3 0

2 −1 1 4 04 3 2 7 0

→ R2 −2R1; R3 −4R1

∼ 1 2 0 −3 0

0 −5 1 10 00 −5 2 19 0

→ R3 −R2

∼

1 2 0 −3 00 −5 1 10 00 0 1 9 0

(3.5)

∴ ρ (A) =ρ (A : Z ) = 3 < 4

System has one-parametric non-trivial (non-zero) solution, that is we are not getting all c1,c2,c3 and c4 arezero. So x4 depends on x1, x2, x3.â In this case to find relation among them, solving (3.5) for c1,c2,c3,c4, we get

c1 = 13t

5, c2 = t

5, c3 =−9t , c4 = t , t ∈R

Substitute in linear combination of (3.4),

13t

5x1,+ t

5x2 −9t x3 + t x4 = 0, t ∈R ⇒ 13x1,+x2 −45x3 +5x4 = 0

Exercise 3.1

1. Examine for linear dependence or independence the following system of vectors. If dependence, findrelation among them: (1-4)

a. −→x1 = (1,−1,1) , −→x2 = (2,1,1) , −→x3 = (3,0,2)

b. −→x1 = (2,2,7,−1) , −→x2 = (3,−1,2,4) , −→x3 = (1,1,3,1)

c. −→x1 = (3,1,−4) , −→x2 = (2,2,−3) , −→x3 = (0,−4,1)

d. −→x1 =[

1 2 4]T

, −→x2 =[

3 7 10]T

2. Which pair of the following vectors are orthogonal:

x = (5,4,1) , y = (3,−4,1) , z = (1,−2,3)

[Hint: For orthogonal vector, dot product is zero.]

3. Find the constant k such that the vectors (2,3k,−4,1,5) and (6,−1,3,7,2k) are orthogonal.

4. Discuss and find the relation of linear dependence among the row vectors of the matrix, 1 1 −1 11 −1 2 −13 1 0 1

5. If a and b are unit vectors such that a+2b and 5a−4b are perpendicular to each other, then find angle

between a and b.

6. Pythagoras4 theorem: For the orthogonal vectors x, y ∈Rn , prove that∥∥x + y∥∥2 = ∥∥x

∥∥2 +∥∥ y∥∥2.

4Pythagoras, Greek; 570-495 BC





7. Parallelogram Law: For any vectors x, y ∈Rn , prove that∥∥x + y∥∥2 +∥∥x − y

∥∥2 = 2∥∥x

∥∥2 +2∥∥ y

∥∥2.

8. For any vectors x, y ∈Rn , prove that ∣∣ ∥∥x∥∥−∥∥ y

∥∥ ∣∣É ∥∥x − y∥∥ .

Answers

1. a. L.D., −→x1 +−→x2 =−→x3 b. L.I. c. L.D., 2−→x1 = 3−→x2 +−→x3 d. L.I. 2. x, y and y , z 3. −1

4. L.D., 2R1 +R2 = R3 5. 60

E E E

Powered by





Contact: (+91) 9427 80 9779





Chapter 4Vector Space

4.1 Field

A non empty set F is said to be a field if it satisfies the following properties:

i. ∀x, y ∈ F, x + y ∈ F and x · y ∈ F. [Closed under addition and multiplication]

ii. ∀x ∈ F ∃ −x ∈ F and ∀x 6= 0 ∈ F ∃ 1

x∈ F . [Existence of additive and multiplicative inverse]

iii. 0,1 ∈ F. [Existence of additive and multiplicative identities]

e. g.

â The set of rational numbers,Q and the set of real numbers, R are real fields.

â The set of complex numbers, C is a complex field.

â The set of natural numbers,N is not a field because 0 ∉N and ∀x ∈N,−x,1

x∉N.

â The set of integers, Z is also not a field because ∀x ∈Z,1

x∉Z.

4.2 Vector Space

A non empty set V is said to be a vector space or a linear space over the field F if there exist two maps+++ : V ×V →V as +++(

u, v)= u+++v , called vector addition (VA), and ··· : F ×V →V as ···(α, v

)=α ···u, called scalarmultiplication (SM), satisfying the following properties:

∀u, v , w ∈V and α,β ∈ F

1. u + v ∈V. [Closed under VA]

2. u + (v +w

)= (u + v

)+w . [Associative law for VA]

3. u + v = v +u. [Commutative property for VA]

4. There exist an element 0 ∈V , such that, ∀u ∈Vu +0 = 0+u = u. [Additive identity]

5. ∀u ∈V there exist an element −u ∈V such that,u + (−u

)= (−u)+u = 0. [Additive inverse]

6. α ·u ∈V. [Closed under SM]

7. α · (u + v)=α ·u +α · v .

26




8.(α+β) ·u =α ·u +β ·u.

9.(αβ

) ·u =α(β ·u

).

10. 1 ·u = u.

Remark:

1. The elements of V are called vectors even though they are any objects like matrices, polynomials,functions, n − tuples etc. The vector space is also known as Abstract vector space.

2. Sometimes vector addition and scalar multiplication are also denoted by ⊕ and ∗ respectively.

3. Instead of field F if we take R, the set of real numbers then V is called the real vector space or reallinear space or real vector linear space overR. Generally we consider always F =R unless given.

4. The scalar multiplication can simply denoted by(αu

)instead of

(α ·u

)4.3 Some Standard Vector Spaces

1. The n dimensional space Rn is a vector space over R under usual addition and scalar multiplicationin Rn .

Let x = (x1, x2.....x3) , y = (y1, y2.....y3

) ∈ Rn and α ∈ R then the usual vector addition and scalar multi-plication in Rn are defined as

x + y = (x1, x2, .....xn)+ (y1, y2, .....yn

)= (x1 + y1, x2 + y2, .....xn + yn

)and

α · x =α (x1, x2, .....xn) = (αx1,αx2, .....αxn)

2. The set of all (2×2) matrices that is M22 with real entries, is a vector space under the matrix additionand matrix scalar multiplication over R.

Here, M22 =[

a bc d

]: a,b,c,d ∈R

Let u =

[u1 u2

u3 u4

], v =

[v1 v2

v3 v4

]∈ M22 and α ∈ R then the matrix addition and scalar multiplica-

tion in M22 are defined as

u + v =[

u1 + v1 u2 + v2

u3 + v3 u4 + v4

]and

α ·u =[αu1 αu2

αu3 αu4

]

3. The set of all polynomial with real coefficients, of degree É n that is Pn (R) is a vector space over R.

Here, Pn (R) = p : p = p (x)/deg p (x) É n

Let p, q ∈Pn and α ∈R then

p = p (x) = a0 +a1x +a2x2 + .....+an xn , q = q (x) = b0 +b1x +b2x2 + .....+bn xn where ai ,bi ∈R.

The vector addition and scalar multiplication in Pn(R) are defined as

p +q = (a0 +b0)+ (a1 +b1) x + (a2 +b2) x2 + .....+ (an +bn) xn

andα ·p = (αa0)+ (αa1) x + (αa2) x2 + .....+ (αan) xn





4. The class of all functions f :R→R that is A is vector spaces under the functions addition and functionscalar multiplication over R.

Let f , g ∈A and α ∈R the vector addition and scalar multiplication in A are defined as(f + g

)x = f (x)+ g (x)

and (α · f

)x =α f (x) ∀x ∈R

Illustration 4.1 Show that the set of all pairs, of real numbers of the form(1, y

)with the operation

(1, y

)+(1, y ′)= (

1, y + y ′) and k(1,k y

), where k ∈R is a vector space.

Solution: Let V = (1, y

): y ∈R

. In order to prove V is a vector space, we have to show that all ten condi-tions for vector space listed in definition of vector space are satisfied.

Let u, v , w ∈V and k,m ∈R ∴ u = (1, x) , v = (1, y

), w = (1, z) for some x, y, z ∈R

1. u + v = (1, x)+ (1, y

)= (

1, x + y) ∈V

[∵ x, y ∈R ⇒ x + y ∈R]

∴ V is closed under vector addition.

2. u + (v +w

)= (1, x)+ [(1, y

)+ (1, z)]

= (1, x)+ [(1, y + z

)]= (1, x)+ (

1, y + z)

= (1, x + y + z

)(u + v

)+w = [(1, x)+ (

1, y)]+ (1, z)

= [(1, x + y

)]+ (1, z)

= (1, x + y

)+ (1, z)

= (1, x + y + z

)u + (

v +w)= (

u + v)+w

∴ Vector addition is associative in V .

3. u + v = (1, x)+ (1, y

)= (

1, x + y)

= (1, y +x

) [∵ x + y = y +x for x, y ∈R]

= (1, y

)+ (1, x)

u + v = v +u

∴ Vector addition is commutative in V .

4. For additive identity, we need to find an element say 0 = (1,θ) ∈ V for some θ ∈ R such that, ∀u ∈V , u +0 = 0+u = u, That is

(1, x)+ (1,θ) = (1,θ)+ (1, x) = (1, x) ⇒ (1, x +θ) = (1,θ+x) = (1, x)

Observe that, above condition holds if θ = 0. Hence 0 = (1,0) ∈V is the additive identity.

∴ Additive identity exist for vector addition in V.

5. For additive inverse, we need to find an element say −u = (1,λ) ∈ V for some λ ∈ R such that ∀u ∈V , u + (−u

)= (−u)+u = 0. That is

(1, x)+ (1,λ) = (1,λ)+ (1, x) = (1,0) ⇒ (1, x +λ) = (1,λ+x) = (1,0)

Observe that, above condition holds if λ = −x. Hence −u = (1,−x) ∈ V is the additive inverse of u =(1, x).

∴ Additive inverse exist for vector addition in V.





6. ku = k (1, x) = (1,kx) ∈V [∵ k, x ∈R ⇒ kx ∈R]

∴ V is closed under scalar multiplication.

7. k(u + v

)= k[(1, x)+ (

1, y)]

= k[(

1, x + y)]

= k(1, x + y

)= (

1,kx +k y) [

∵ By definition of SM]

ku +kv = k (1, x)+k(1, y

)= (1,kx)+ (

1,k y)

= (1,kx +k y

) [∵ By definition of VA

]∴ k

(u + v

)= ku +kv

8. (k +m)u = (k +m) (1, x)

= [1, (k +m) x]

= (1,kx +mx)

ku +mu = k (1, x)+m (1, x)

= (1,kx)+ (1,mx)

= (1,kx +mx)

∴ (k +m)u = ku +mu

9. (km)u = (km) (1, x)

= [1, (km) x]

= (1,kmx)

k(mu

)= k [m (1, x)]

= k [(1,mx)]

= k (1,mx)

= (1,kmx)

∴ (km)u = k(mu

)10. 1u = 1(1, x)

= (1,1x)

= (1, x) [∵ 1x = x]

∴ 1u = u

Thus, all ten conditions for vector space are hold true for given vector addition and scalar multiplication inV . Therefor V is a vector space.

Illustration 4.2 Check whether the set V = (x, y

): x, y ∈R

, under the addition(x1, y1

)⊕(x2, y2

)= (x1 +x2, y1 + y2

)and multiplication α∗ (

x, y)= (

α2x,α2 y), is a vector space over the field R or not?

Solution: Given that V = R2 and the defined addition is the usual vector addition of R2, hence all fiveconditions for vector space are satisfied evidently. So it is sufficent to check remaining five conditions forscalar multiplication.Let u = (

x, y) ∈V and α,β ∈R.

1. αu =α(x, y

)= (

α2x,α2 y) ∈V

[∵ α, x, y ∈R ⇒ α2x,α2 y ∈R]

∴ V is closed under scalar multiplication.





2.(α+β)

u = (α+β)(

x, y)

=[(α+β)2x,

(α+β)2 y

] [∵ By definition of SM

]= [(

α2 +2αβ+β2)x,(α2 +2αβ+β2) y

]= (

α2x +2αβx +β2x,α2 y +2αβy +β2 y)

αu +βu =α(x, y

)+β(x, y

)= (

α2x,α2 y)+ (

β2x,β2 y)

= (α2x +β2x,α2 y +β2 y

)∴

(α+β)

u 6=αu +βu

That is scalar multiplication is not distributive over scalar addition. Hence V is not a vector space. (Readercan verify that all other remaining conditions for scalar multiplications are hold)

4.4 Subspace

A non empty subset W of the vector space V over R, is said to be a subspace of V if, W itself vector spaceover R, udder the same vector addition and scalar multiplication of V .

Theorem 4.1 A non empty subset W of a vector space V over R, is subspace of V if and only if,

i. ∀u, v ∈W then u + v ∈W.

ii. ∀u ∈W,α ∈R then αu ∈W.

That is, W should be closed under vector addition and scalar multiplication.

Note: Every vector spcae V has two precise subspace like, singleton set 0 and vector space V itself. Thesesubspces are called trivial subspace.

Illustration 4.3 Check whether the following subsets W of vector space V are subspaces or not?

a. W = (x,3x,2x) : x ∈R ;V =R3. b. W = (x, y, z

): x2 + y2 + z2 É 1

;V =R3.

c. W = The set of all points lying on the line passing through the origin and V =R2.

Solution: In order to check subspace, first of all we show that given set W is a non empty subset of V (thatis to show atleast one element exist in W ), and then we check two conditions of Theorem 4.1.

a. Here W = (x,3x,2x) : x ∈R ;V =R3.

Obviously W ⊂V , and for 0 ∈R, 0 = (0,0,0) = (0,3(0) ,2(0)) ∈W . Therefor, W is non empty.

Let u, v ∈W,α ∈R, therefor u = (x,3x,2x) , v = (y,3y,2y

)for some x, y ∈R.

i. u + v = (x,3x,2x)+ (y,3y,2y

)=

(x + y,3x +3y,2x +2y

)=

(x + y,3

(x + y

),2

(x + y

)) ∈W[∵ x, y ∈R ⇒ x + y ∈R]

ii. αu =α (x,3x,2x)

= (αx,3αx,2αx)

= (αx,3(αx) ,2(αx)) ∈W [∵α, x ∈R ⇒ αx ∈R]

∴ W is closed under vector addition and scalar multiplication.

∴ W is a subspace of V .

b. Here W = (x, y, z

): x2 + y2 + z2 É 1

;V =R3.

Obviously W ⊂V , and for 0 = (0,0,0) ∈R3,02 +02 +02 É 1. Therefor, 0 ∈W , so W is non empty.

Let u = (x1, y1, z1

), v = (

x2, y2, z2) ∈W ⇒ x2

1 + y21 + z2

1 É 1, x22 + y2

2 + z22 É 1

[By definition of W

]LAVC (GTU-2110015) B.E. Semester II




i. u + v = (x1 +x2, y1 + y2, z1 + z2

).

Now,

(x1 +x2)2 + (y1 + y2

)2 + (z1 + z2)2 = (x2

1 +2x1x2 +x22

)+ (y2

1 +2y1 y2 + y22

)+ (z2

1 +2z1z2 + z22

)= (

x21 + y2

1 + z21

)+ (x2

2 + y22 + z2

2

)+ (2x1x2 +2y1 y2 +2z1z2

)É 1+1+ (

2x1x2 +2y1 y2 +2z1z2)

É 2+ (2x1x2 +2y1 y2 +2z1z2

)Ð 1 (always)

∴ u + v does not satisfy condition of W . ∴ u + v ∉W

∴ W is not closed under vector additon. hence W is not subspace of V.

Note: Observe that, geometrically W represent the interior of the unit sphere.(i.e. x2 + y2 + z2 É 1

).

Hence the interior of the unit sphere is not a subspace of the whole space.

c. Equation of line passing through origin is given by y = mx,m ∈R.

∴ W = (x, y

): y = mx,m ∈R

, V =R2.

Obviously W ⊂V , and 0 = (0,0) ∈W . Therefor, W is non empty.

Let u = (x1, y1

), v = (

x2, y2) ∈W and α ∈R ⇒ y1 = mx1, y2 = mx2, m ∈∈R.

i. u + v = (x1, y1

)+ (x2, y2

)= (x1 +x2, y1 + y2

)y1 + y2 = mx1 +mx2 = m (x1 +x2) ⇒ u + v ∈W

ii. αu =α(x1, y1

)= (αx1,αy1

)αy1 =α (mx1)m (αx1) ⇒ αu ∈W

∴ W is closed under vector addition and scalar multiplication.

∴ W is a subspace of V .

Exercise 4.1

1. Check whether the following sets are vector space over the field R or not?

a. V = x > 0 : x ∈R where x + y = x y and αx = xα.

b. V = (x, y

): x, y ∈N

where(x1, y1

)+ (x2, y2

)= (x1 +x2, y1 + y2

)and α

(x, y

)= (αx,αy

).

c. V =R2 where(x1, y1

)⊕ (x2, y2

)= (x1 +x2 +1, y1 + y2 +1

)and k ¯ (

x, y)= (

kx,k y)

d. V = p ∈P2 : p (0) = 1

with usual operations.

e. V =R3 with usual vector addition and scalar multiplication defined by k(x, y, z

)= (0,0,kz) .

f. V =R2 where (u1,u2)⊕(v1, v2) = (u1 + v1 −2,u2 + v2 −3) andα¯(u1,u2) = (αu1 +2α−2,αu2 −3α+3)

2. Explain why the set of all 2-by-2 matrices with rational entries is not a real vector space?

[Hint: Not closed under scalar multiplication.]

3. Check whether the following subsets W of vector space V , under usual operations, are subspace ornot?

a. W = (x,0,0) : x, y ∈R

; V =R3. b. W = (x, y, z

): x2 + y2 + z2 = 1

;V =R3.

c. W =

a 00 bc 0

: a,b,c ∈R ;V =M32. d. W = (

x, x2,0)

: x ∈R; V =R3.

e. W = The set of all 2-by-2 symmetric matrices; V =M22.

f. W = (x, y, z

) ∈R3 : x + y + z = 1

; V =R3.

g. W = f ∈V : f (0) = 0

; and V = the set of all real valued functions.





h. W = The interior of the unit circle and V =R2.

i. W = The set of all 2-by-2 skew-symmetric matrices and V =M22. .

j. W = The set of all the lines not passing through origin and V =R2.

k. W =The plane does not passing through origin and V =R3.

l. W = f ∈V : f (0) = 1

; V =A .

4. Prove that the circular cylinder generated by unit circle is not a subspace of the space, under usual VAand SM.

[Hint: W = (x, y, z

) ∈R3 : x2 + y2 = 1, z ∈R]

5. Prove that if W1 and W2 are subspaces of the vector space V then W! ∪W2 is also subspace of V . ButW! ∩W2 may not be sub space of V .

Answers

1. a. Yes, all other no. 3. a, c, e, g, i are subspace, all other are not subspace.

E E E

4.5 Linear Combination and Span

â Linear combination of the vectors v1, v2, v3.....vn of a vector space V , is defined as

c1v1 + c2v2,+c3v3 + .....+ cn vn , c1,c2,c3.....cn ∈R

â Set of all linear combinations of vectors of W = v1, v2, v3.....vn

is called span of W and is denoted by

spanW , that is

spanW = span

v1, v2, v3.....vn=

c1v1 + c2v2 + c3v3 + .....+ vn : ci ∈R

e. g. v1 = (1,−1,2) , v2 = (3,2,1) then

span

v1, v2= c1 (1,−1,2)+ c2 (3,2,1) : c1,c2 ∈R = (c1 +3c2,−c1 +2c2,2c1 + c2) : c1,c2 ∈R

â We can say that a w vector of a vector space V is a linear combination of the vectors v1, v2, v3.....vn ofa vector space V , if there exist scalars c1,c2,c3.....cn ∈R such that,

w = c1v1 + c2v2,+c3v3 + .....+ cn vn

â If all the vectors of V are expressed as a linear combustion of the vectors v1, v2, v3.....vn then we cansay that set

v1, v2, v3.....vn

span V and is denoted by span

v1, v2, v3.....vn

=V.

â Let v1, v2, v3.....vn be the vectors of a vector space V and if vn+1 ∈ span

v1, v2, v3.....vn

, then

span

v1, v2, v3.....vn= span

v1, v2, v3.....vn , vn+1

â For the subset W =

v1, v2, v3.....vn

of a vector space V , spanW is subspace of V .

4.6 Linearly Independent Vectors (LI)

The vectors v1, v2, v3.....vn ∈V are said to be linearly independent if whenever,

c1v1 + c2v2 + c3v3 + .....+ cn vn = 0 ⇒ c1 = c2 = c3 = ..... = cn = 0.





4.7 Linearly Dependent Vectors (LD)

The vectors v1, v2, v3.....vn ∈V are said to be linearly dependent if whenever,

c1v1 + c2v2 + c3v3 + .....+ cn vn = 0 ⇒ c1,c2,c3, .....cn not all zero

â In general the set

v1, v2, v3.....vn ⊆ V is linearly dependent if and only if at least one vector can

always be expressed as a linear combustion of rest of the vectors.

â In particular, two vectors u and v are linearly dependent if and only if u = kv for some k ∈R.e. g.

i. u = (2,−1,3) , v = (6,−3,9) ∈R3 are linearly dependent because v = 3u.

ii. sin2x and sin x cos x are also linearly dependent because sin2x = 2sin x cos x.

â A finite set of vectors that contains a zero vector, 0 is always linearly dependent .e. g. (3,−1,2) , (1,2,−4) , (0,0,0) is linearly dependent set because it contains zero vector.

â A singleton set (set containing only one vector) is linearly dependent if and only if it contains a zerovector. That is 0.

e. g. (1,0,−1) is linearly independent where as (0,0,0) is linearly dependent .

4.8 Wronskian1

The Wronskian of the functions u, v or u, v, w is defined as a determinant

W =∣∣∣∣ u v

u′ v ′∣∣∣∣ or W =

∣∣∣∣∣∣u v wu′ v ′ w ′

u′′ v ′′ w ′′

∣∣∣∣∣∣â The subset of functions u, v or u, v, w of C (set of all continuous functions) is LD OR LI according tothe corresponding Wroskian of the functions W = 0 OR W 6= 0.

Illustration 4.4 Show that vector w = (9,2,7) is a linear combination of u = (1,2,−1) and v = (6,4,2) in R3.

Solution: To show ←−w is a linear combination of ←−u and ←−v , we have to find constants c1 and c2 such that

c1u + c2v = w ⇒ c1 (1,2,−1)+ c2 (6,4,2) = (9,2,7) (4.1)

To solve non-homogeneous system (4.1), consider the augmented matrix [A : B ], can be obtained by puttingthe vectors in columns as

[A : B ] = 1 6 9

2 4 2−1 2 7

→ R2 −2R1; R3 +R1

∼ 1 6 9

0 −8 −160 8 16

→ R3 +R2

∼ 1 6 9

0 −8 −160 0 0

∴ ρ (A) =ρ (A : B) = 2 (number of unknowns)

∴ System has unique solution and is given by back substitution as

c1 +6c2 = 9, −8c2 =−16 ⇒ c1 =−3,c2 = 2

∴ From (4.1), w =−3u +2v

1Jòzef Maria Hoene-Wronski; Polishsh, 1776-1853.





Illustration 4.5 Show that span(1,0,1) , (−1,2,3) , (0,1,−1) =R3

Solution: To show the span, we have to show that every vector u = (x, y, z

) ∈R can be expressed as a linearcombination of the given vector. That is there should exist c1,c2,c3 such that

c1 (1,0,1)+ c2 (−1,2,3)+ c3 (0,1,−1) = (x, y, z

)(4.2)

Consider the augmented matrix of system (4.2)

[A : B ] = 1 −1 0 x

0 2 1 y1 3 −1 z

→ R3 −R1

∼ 1 −1 0 x

0 2 1 y0 −4 −1 z −x

→ R3 +2R2

∼ 1 −1 0 x

0 2 1 y0 0 1 z −x +2y

∴ ρ (A) =ρ (A : B) = 3 (no of unknowns.)

∴ System is consistent and has unique solution.∴ There exist c1,c2,c3, satisfying (4.2). Hence span(1,0,1) , (−1,2,3) , (0,1,−1) =R3

Note: To find linear combination for given vector of R3, solve above system by back substitution, we get

c1 = 1

2

(3x − y − z

), c2 = 1

2

(x − y − z

), c3 =−x +2y + z

Hence from (4.2)

3x − y − z

2(1,0,1)+ x − y − z

2(−1,2,3)+ (−x +2y + z

)(0,1,−1) = (

x, y, z)

e. g. If u = (1,1,1) ⇒ c1 = 1

2,c2 =−1

2,c3 = 2 ∴ (1,1,1) = 1

2(1,0,1)− 1

2(−1,2,3)+2(0,1,−1)

Illustration 4.6 Examine the following vectors for LI or LD:

a. 1− t + t 3,−2+3t + t 2 +2t 3,1+ t 2 +5t 3 in P3. b.[

1 −11 1

],

[ −2 31 2

],

[1 01 0

]in M22.

c. cos2x, sin2x in R .

Solution:a. Consider the linear combination of given vectors of P3 :

c1(

1− t + t 3)+ c2(−2+3t + t 2 +2t 3)+ c3

(1+ t 2 +5t 3)= 0 = 0+0 · t +0 · t 2 +0 · t 3 (4.3)

The augmented matrix for corresponding homogeneous system of (4.3) is, (obtained by putting coefficientascending powers of each polynomial in columns)

c1 c2 c3

[A : Z ] =

1 −2 1 0

−1 3 0 00 1 1 01 2 5 0

1t

t 2

t 3





Reducing to row echelon form:

[A : Z ] =

1 −2 1 0

−1 3 0 00 1 1 01 2 5 0

→ R2 +R1; R4 −R1

∼

1 −2 1 00 1 1 00 1 1 00 4 4 0

→ R3 −R2; R4 −4R2

∼

1 −2 1 00 1 1 00 0 0 00 0 0 0

∴ ρ (A) =ρ (A : Z ) = 2 < 3 (number of unknowns)

∴ System (4.3) has non-trivial (non zero) one parametric solution. Hence all c1,c2,c3 can not be zero.Hence given vector are linearly dependent.Note: To find dependent relation, solving above system we get c2 =−c3,c1 =−3c3.∴ From (4.3), −3

(1− t + t 3)− (−2+3t + t 2 +2t 3)+ (

1+ t 2 +5t 3)= 0

b. Consider the linear combination of given vectors of M22 :

c1

[1 −11 1

]+ c2

[ −2 31 2

]+ c3

[1 01 0

]= 0 =

[0 00 0

](4.4)

The augmented matrix for corresponding homogeneous system of (4.4) is, (obtained by putting row-entriesof each matrix in columns)

[A : Z ] =

1 −2 1 0−1 3 0 01 1 1 01 2 0 0

→ R2 +R1; R3 −R1; R4 −R1

∼

1 −2 1 00 1 0 00 −3 1 00 4 −1 0

→ R3 +3R2; R4 −4R2

∼

1 −2 1 00 1 0 00 0 1 00 0 −1 0

→ R4 +R3

∼

1 −2 1 00 1 0 00 0 1 00 0 0 0

→ R4 +R3

∴ ρ (A) =ρ (A : Z ) = 3 (number of unknowns)

∴ System (4.4) has unique trivial (zero) solution, that is c1 = c2 = c3 = 0. Hence given vectors are linearlyindependent.

c. Given functions sin2x and cos2x are linearly in dependent because we can not write one function asa constant multiple of another function. That is sin2x 6= k ·cos2x, k ∈R.Alternate Method:





Let u = sin2x, v = cos2x, then Wronskian’s of u, v is

W =∣∣∣∣ u v

u′ v ′∣∣∣∣= ∣∣∣∣ sin2x cos2x

sin2x −sin2x

∣∣∣∣= sin2x

∣∣∣∣ sin2x cos2x1 −1

∣∣∣∣=−sin2x 6= 0

∴ u and v are linearly independent.

Illustration 4.7 Find the condition on parameter a such that the set (1,−1,1) , (1,0, a) , (−1,−a,0) is lin-early independent.

Solution: Three vectors of R3 are linearly independent if the determinant of vectors is not zero. That is∣∣∣∣∣∣1 1 −1−1 0 −a1 a 0

∣∣∣∣∣∣= a2 6= 0 ⇒ a 6= 0

∴ Given vectors are linearly independent for all a ∈R− 0 .

Exercise 4.2

1. Find the span of the vectors (1,0,0) and (0,0,1).

2. Express (5,−1,9) as a linear combination of v1 = (2,9,0) , v2 = (3,3,4) , v3 = (1,2,1) . [Summer-2016]

3. Is (4,20) is linear combination of the vectors (2,10) and (−3,−15) ?

4. Show that in R4 the vector (1,4,−2,6) is a linear combination of the vectors (1,2,0,4) and (1,1,1,3)where as (2,6,0,9) is not a linear combination of given vectors.

5. Show that the matrices

[1 00 0

],

[1 10 0

],

[1 11 0

],

[1 11 1

]span M22

6. Check the LI or LD for the set x, |x | in C (1,1).

7. If v1, v2, v3 are linearly independent in V then prove that(v1 + v2

),(v2 + v3

),(v3 + v1

)are also linearly

independent.

8. Prove that every subset of LI set is LI and every super set of LD set is LD.

9. Let

v1, v2, .....vn

be LI subset of vector space V over R. If x1, x2, .....xn and

y1, y2, .....yn

be two

subsets of R such thatn∑

i=1xi v i =

n∑i=1

yi v i then prove that xi = yi ∀ 1 É i É n

Answers

1. R2 2. −308v1 +69v2 +179v3 3. Yes, (4,20) = 5(2,10)+2(−3,−15) 6. LD

E E E

4.9 Basis

A subset W = v1, v2, v3.....vn

of a vector space V is said to be the basis of V if,

i. v1, v2, v3.....vn are linearly independent and

ii. W span V i.e. span W =V

â If W is basis for V then we can say that V is generated by W and W is called generator of V .





4.10 Some Standard Basis

1. The standard basis of R2 is (1,0) , (0,1), of R3 is (1,0,0) , (0,1,0) , (0,0,1) and so on.

2. The standard basis of M2×2 is

[1 00 0

],

[0 10 0

],

[0 01 0

],

[0 00 1

],

3. The standard basis of Pn (x) is1, x, x2, x3, ......xn

Note: A vector space can have more than one basis. But every basis number of the vectors is same. (vectorsmay be different)

4.11 Dimension

The number of the vectors in any basis of the vector space V is said to be the dimension of V and is denotedby dimV . â If dimV is finite then V is called finite dimensional vector space.

e. g. dim(Rn)= n, dim(Pn) = (n +1) , dim(M22) = 4.

* Important:

If dim v = n, then

1. Every basis of V has exactly n number of vectors.

2. A subset of less or more than n vectors could not be a basis of V.

3. Set of more than n vectors is always linearly dependent (LD)

4. Set of less than n vectors could not be spanV . (No span)

5. A subset of exactly n vectors, which is either LI or span V is always basis of V .

Illustration 4.8 Show that whether the following sets form basis for given vector space or not? Justify theanswers.

a. (1,2) , (3,−1) for R2. b. (1,1,0) , (−1,0,0) for R3.

c. (1,−1,1) , (−1,2,−2) , (−1,4,−4) for R3. d. (1,0,1) , (1,1,0) , (0,1,1) , (2,1,1) for R3.

e.3+x3,2−x −x2, x +x2 −x3, x +2x2 for P3 (x) .

Solution:

a. We know that dimR2 = 2 and given set contain exactly two vectors. So it is sufficient to check weatherthe set is linearly independent of not.

For two vectors of R2, the determinant of the vectors is

∣∣∣∣ 1 32 −1

∣∣∣∣=−7 6= 0.

∴ Given subset is linearly independent subset of R2, hence it is basis.

b. Given subset has two vector and dimR3 = 3, hence it can not span R3. So it is not basis.

c. Given subset has three vectors and dimR3 = 3. Hence for basis it sufficient to check linearly dependentof given subset.

For three vectors of R3, the determinant of the vectors is

∣∣∣∣∣∣1 −1 −1−1 2 41 −2 −4

∣∣∣∣∣∣= 0.

∴ Given subset is linearly dependent. So it is not basis.





d. Given subset has four vector and dimR3 = 3, hence it linearly dependent subset. So it is not basis.

e. Given subset has four vectors (polynomials) and dimP3 = 4, so it is sufficient to check linearly independence of given polynomials. Consider the linear combination,

c1(3+x3)+ c2

(2−x −x2)+ c3

(x +x2 −x3)+ c4

(x +2x2)= 0

The augmented matrix for above system is

[A : Z ] =

3 2 0 0 00 −1 1 1 00 −1 1 1 01 0 −1 0 0

R1 ↔ R4

∼

1 0 −1 0 00 −1 1 1 00 −1 1 1 03 2 0 0 0

→ R4 −3R1

∼

1 0 −1 0 00 −1 1 1 00 −1 1 1 00 2 3 0 0

→ R3 −R2; R4 +2R2

∼

1 0 −1 0 00 −1 1 1 00 0 0 0 00 0 5 2 0

R3 ↔ R4

∼

1 0 −1 0 00 −1 1 1 00 0 5 2 00 0 0 0 0

∴ ρ (A) =ρ (A : Z ) = 3 < 4 (number of unknowns)

∴ System has one parametric non trivial (non zero) solution, that is not all c1,c2,c3,c4 are zero.∴ Given subset is linearly dependent and hence it is not basis.

Illustration 4.9 Reduce the following set (1,0,0) , (0,1,−1) , (0,4,−3) , (0,2,0) to obtain the basis for thevector space R3.

Solution: We know that dim(R3)= 3 and given set has four vectors. So it is linearly dependent , and not a

basis. To reduce to basis of R3, we have to remove one vector from set which depends on other vectors. Itcan be done as follow:

â Construct a matrix say A, by taking the vectors in column.

â Reduce matrix A to its equivalent row echelon form.

â The vectors corresponding to non pivot columns are linearly dependent vectors, that will be removedfrom original set, and we required basis.

Let v1 = (1,0,0) , v2 = (0,1,−1) , v3 = (0,4,−3) , v4 = (0,2,). Matrix of vectors,

v1 v2 v3 v4

A = 1 0 0 0

0 1 4 20 −1 −3 0

→ R3 +R2





∼

1 0 0 00 1 4 20 0 1 2

Since the fourth column in echelon form is non pivoting, on removing corresponding v4 = (0,2,0) fromgiven set we get required basis. That is (1,0,0) , (0,1,−1) , (0,4,−3) ,

Illustration 4.10 Find the basis for the solution space of the equation AX = 0, where

A =

−1 0 1 2−1 1 0 −10 −1 1 31 −2 1 4

Solution: To find basis for solution space of the equation AX = 0, first of all we obtain solution of the givenhomogeneous system. Consider the augmented matrix for given equation is given by

x1 x2 x3 x4

[A : Z ] =

−1 0 1 2 0−1 1 0 −1 0

0 −1 1 3 01 −2 1 4 0

→ R2 −R1; R4 +R1

∼

−1 0 1 2 0

0 0 −1 −3 00 −1 1 3 00 −2 2 6 0

R2 ↔ R3

∼

−1 0 1 2 0

0 −1 1 3 00 0 −1 −3 00 −2 2 6 0

→ R4 −2R2

∼

−1 0 1 2 0

0 −1 1 3 00 0 −1 −3 00 0 0 0 0

∴ ρ (A) =ρ (A : Z ) = 3 < 4

∴ System has non trivial one parametric solution which is obtained by assuming parameter t ∈ R to freevariable x4, and is given by

x1 =−t , x2 =−2t x3 =−3t , x4 = t , t ∈RNow solution space is defined as set of all possible solutions, that is

W =

X : AX = 0

= (x1, x2, x3, x4) : x1 =−t , x2 =−2t , x3 =−3t , x4 = t , t ∈R

= (−t ,−2t ,3t , t ) : t ∈R

= t (−1,−2,−3,1) : t ∈R

=Linear combination of the vector (−1,−2,−3,1) .

=span(−1,−2,−3,1)

∴ Solution space is spaned by the set (−1,−2,−3,1) and set is also linearly independent because it on-tained only one non zero vector.∴ (−1,−2,−3,1) is required basis for solution space.

Note that, dimension of the solution space is 1, and it is same as number of non pivot column of rowechelon form of A.





Illustration 4.11 Find the basis for the plane x +2y +3z = 0 in R3.

Solution: Let W be the given plane x +2y +3z = 0 in R3.

∴ W =(x, y, z

): x +2y +3z = 0, x, y, z ∈R

=(−y −3z, y, z)

: y, z ∈R [∵ x +2y +3z = 0 ⇒ x =−y −3z

]=(−y, y,0

)+ (−3z,0, z) : y, z ∈R [Separating y and z

]=

y (−1,1,0)+ z (−3,0,1) : y, z ∈R∴ W =span(−1,1,0) , (−3,0,1)

∴ W is spanned by linearly independent set (−1,1,0) , (−3,0,1).∴ required basis for plane x +2y +3z = 0 is (−1,1,0) , (−3,0,1) and dimension 0f plane is 2.

Exercise 4.3

1. Which of the following sets of vectors are basis ?

a. 1−3x +2x2,1+x +4x2,1−7x for P2

b.[

1 21 −2

],

[0 −1−1 0

],

[0 23 1

],

[0 0−1 2

]for M22 [Winter-2015]

2. Let V be the space spanned by v1 = cos2x, v2 = sin2x, v3 = cos2x then show that S = v1, v2, v3

is

not basis for V .

[Hint: cos2x = cos2x − sin2x]

3. For what real values of λ do the following vectors form a basis for R3 ?

v1 =(λ,−1

2,−1

2

), v2 =

(−1

2,λ,−1

2

), v3 =

(−1

2,−1

2,λ

)[Hint: Take determinant of vectors = 0]

4. Reduce the set (0,1,1) , (1,1,−1) , (3,1,−3) , (1,2,0) to basis for R3.

5. Extend the set (1,1,1,1) , (1,2,1,2) to a basis for R4. [Winter-2012]

[Hint: Take union with standard basis of R4 and then reduce new set of six vectors.]

6. Extend the set1, x2 to a basis for P4.

7. Reduce the following set to obtain the basis for the vector space P2:p0 = 2, p1 =−4x, p2 = x2 +x +1, p3 = 2x +7, p4 = 5x2 −1.

8. In each part, determine whether the three vectors lie in a plane (linearly independent ) or on the sameline (linearly dependent ):

a. v1 = (2,−2,0) , v2 = (6,1,4) , v3 = (2,0,−4) b. v1 = (−6,7,2) , v2 = (3,2,4) , v3 = (4,−1,2)

9. Show that M23 has dimension 6.

[Hint: Construct standard basis for M23. ]

10. Determine a basis for and the dimension of the solution space of the following homogeneous system:

a. 2x1 +2x2 −x3 +x5 = 0

−x1 −x2 +2x3 −3x4 +x5 = 0

x1 +x2 −2x3 −x5 = 0

x3 +x4 +x5 = 0 [Winter-2017]

b. x + y + z = 0

3x +2y −2z = 0

4x +3y + z = 0

6x +5y + z = 0

11. Determine basis for the following subspace of R3 :





a. The line x = 2t , y =−t , z = 4t .

b. All the vectors of the form (a,b,c) for which b = a + c.

c. The subspace(

x, y, z, w) ∈R4 : x + y − z = y + z = 0

.

12. Is

v1, v2

form basis for H ? Where v1 = 1

00

, v2 = 0

10

, H =

xx0

: x ∈R .

13. Under what condition is a set with one vector linearly independent ?

14. Show that every set with more then three vectors from P2 is linearly dependent.

15. Prove that the space spanned by two vectors in R3 is a line through the origin, a plane through theorigin or the origin itself.

16. Use proportional identities, where required, to check which of following sets of vectors in F (−∞,∞)are linearly dependent.

a. 6, 3sin2x, 2cos2x b. x, cos x

c. 1, sin x, sin2x d. (3−x)2, x2 −6x, 5

17. Given two linearly independent vectors (1,0,1,0) and (0,−1,1,0) ofR4, find a basis for R4 that includesthese two vectors.

18. Determine whether the vectors v2 = (1,2,−1) , v3 = (−3,1,0) , v4 = (2,11,−5) forms a basis for R3 or not? If not choose, construct a basis of R3 consisting the vectors out of the given vectors.

Answers

1. a. No b. Yes 3. λ ∈R−−1

2,1

4. (0,1,1) , (1,1,−1) , (3,1,−3)

5. (1,1,1,1) , (1,2,1,2) , (1,0,0,0) , (0,1,0,0) 6.1, x, x2, x3, x4 7.

p0, p1, p2

8. a. in plane. b. on line 10. a. (−1,1,0,0) , (−1,0,−1,0,1) ,dim = 2 b. Null space, dim= 0

11. a. (2,−1,4) b. (1,1,0) , (0,1,1) c. (2,−1,1,0) , (0,0,0,1) ,dim = 2 12. Yes

16. a. d. LD, b. LI 17. (1,0,1,0) , (0,−1,1,0) , (1,0,0,0) , (0,0,0,1)

18. not basis, (1,2,−1) , (−3,1,0) , (1,0,0)

E E E

4.12 Ordered Basis and Coordinate Vector

â An ordered basis is a basis S = v1, v2, v3.....vn

along with the ordering of its vectors.

â Let S = v1, v2, v3.....vn

be the ordered basis for the vector space V and u ∈V .

If u = c1v1+c2v2+c3v3+.....+cn vn , then the coefficients c1,c2,c3, .....cn are called coordinate of vectoru with respect to the basis S.

â The corresponding vector (c1,c2,c3, .....cn) of Rn is called coordinate and is denoted by(u

)S . That is(

u)

S = (c1,c2,c3, .....cn)





4.13 Translation Matrix (Change of Basis Matrix)

â Let S = v1, v2, v3.....vn

and T =

w1, w2, w3.....wn

are two basis for the vector space V.

â The translation matrix from basis T to S is defined as,

P = [[w1

]S ,

[w2

]S ,

[w3

]S .....

[wn

]S

]where i th column of matrix P is the coordinate matrix of wi relative to the basis S

â The coordinate matrix of u relative to S,[u

]S and relative to T ,

[u

]T are related by equation,[

u]

S = P[u

]T

* Important:

1. Sometimes the translation matrix from T to S is denoted by PT→S .

2. Here P is always invertible and P−1 be a translation matrix from S to T , that is P−1S→T

. Hence

PT→S ⇔ P−1S→T

Illustration 4.12 If u = (10,5,0) ∈R3, Find the coordinate vector for u relative to,

a. The standard basis for R3.

b. The basis T = (1,−1,1) , (0,1,2) , (3,0,−1)

Solution:a. The standard basis for R3 is S = (1,0,0) , (0,1,0) , (0,0,1) =

e1,e2,e3

(Say). To find coordinate vectorrelative to standard basis S, we represent u = (10,5,0) as a linear combination of vectors S.

u = (10,5,0)

= (10,0,0)+ (0,5,0)+ (0,0,0)

=10(1,0,0)+5(0,1,0)+0(0,1,0)

u =10 ·e1 +5 ·e2 +0 ·e3

∴(u

)S = (10,5,0) Required coordinate vector relative to standard basis.

â It is worth to note that, for any vector of R3 (in fact of Rn ), given vector it self represent a coordinatevector relative to standard basis.

b. For coordinate vector relative to basis T = (1,−1,1) , (0,1,2) , (3,0,−1) = v1, v2, v3

(say), we represent

u as a linear combination of vectors of T . Consider,

c1v1 + c2v2 + c3v3 = u ⇒ c1 (1,−1,1)+ c2 (0,1,2)+ c3 (3,0,−1) = (10,5,0) (4.5)

To solve non-homogeneous system (4.5), consider augmented matrix: (Put coefficient vectors in column)

c1 c2 c3

[A : B ] = 1 0 3 10

−1 1 0 51 2 −1 0

→ R2 +R1; R3 −R1

∼ 1 0 3 10

0 1 3 150 2 −4 −10

→ R3 −2R1





∼ 1 0 3 10

0 1 3 150 0 −10 −40

Back substitution ⇒ c1 =−2, c2 = 3, c3 = 4

∴ Required coordinate vector relative to basis T is(u

)T = (c1,c2,c3) = (−2,3,4)

Illustration 4.13 Determine the coordinate vector of p = 4−2x+3x2 relative to the basis B = 2,−4x,5x2 −1

for P2.

Solution: Consider the lineal combination,

c1p1 + c2p2 + c3p3 = p ⇒ c1 (2)+ c2 (−4x)+ c3(5x2 −1

)= 4−3x +4x2 (4.6)

To solve system (4.6), consider augmented matrix, (putting coefficients of polynomials in column)

c1 c2 c3

[A : Z ] = 2 0 −1 4

0 −4 0 −20 0 5 3

Observe that above matrix is already in echelon form, so making back substitution we get c1 = 23

10,c2 =

1

2,c3 = 3

5. Hence, Required coordinate is

(p

)B =

(23

10,

1

2,

3

5

).

Illustration 4.14 Consider the standard basis for R3 i.e. S = e1,e2,e3

and another basis T =

v1, v2, v3

where v1 = (1,−1,1) , v2 = (0,1,2) , v3 = (3,0,−1) of R3

a. Find the translation matrix P from T to S and Q from S to T.

b. Compute[u

]S , given that

[u

]T = (9,−1,−8)

c. Compute[u

]T , given that

[u

]S = (−6,7,2)

Solution:a. By definition of translation matrix P from T to S, we have

P = [[v1

]S

[v2

]S

[v3

]S

](4.7)

where[v1

]S ,

[v2

]S ,

[v3

]S are column matrix of coordinate vectors, of v1, v2, v3 relative to standard basis S

respectively.Also know that the coordinate vector of any vector of R3 relative to standard basis is vector itself (See

Illustration 4.12). Hence

[v1

]S =

1−1

1

,[v1

]S =

012

,[v1

]S =

30−1

∴ From (4.7), the translation matrix P from T to S is, P = 1 0 3

−1 1 01 2 −1

Also the translation matrix Q from S to T is given by

Q = [[e1

]T

[e2

]T

[e3

]T

], e1 = (1,0,0) ,e2 = (0,1,0) ,e3 = (0,0,1) (4.8)

where[e1

]T ,

[e2

]T ,

[e3

]T are column matrix of coordinate vectors, of e1,e2, ve3 relative to the basis T re-

spectively.





â To find these coordinate vectors, first we obtain coordinate for any vector say u = (a,b,c) ∈ R3 relativeto basis T. Let

c1v1 + c2v2 + c3v3 = u ⇒ c1 (1,−1,1)+ c2 (0,1,2)+ c3 (3,0,−1) = (a,b,c) (4.9)

Augmented matrix os system (4.9) is

c1 c2 c3

[A : B ] = 1 0 3 a

−1 1 0 b1 2 −1 c

→ R2 +R1; R3 −R1

∼ 1 0 3 a

0 1 3 b +a0 2 −4 c −a

→ R3 −2R2

∼ 1 0 3 a

0 1 3 b +a0 0 −10 c −3a −2b

By back substitution, we get

c1 = 1

10(a −6b +3c) , c2 = 1

10(a +4b +3c) c3 = 1

10(3a +2b − c)

∴[u

]T = [(a,b,c)]T =

c1

c2

c3

= 1

10

a −6b +3ca +4b +3c3a +2b − c

⇒ [e1

]T = [(1,0,0)]T = 1

10

113

,[e2

]T = [(0,1,0)]T = 1

10

−642

,[e3

]T = [(0,0,1)]T = 1

10

33−1

∴ From (4.8), required translation matrix Q from S to T is, Q = 1

10

1 −6 31 4 33 2 −1

Note: We know that if P is a translation matrix from T to S then P−1 is a translation matrix from S to T .

Hence if P is known, then Q = P−1 = 1

10

1 −6 31 4 33 2 −1

(Verify !)

b. To find[u

]S using

[u

]T = (9,−1,−8). That is to convert T coordinate into S coordinate. Hence we

use the translation matrix T to S, that is P, given by the relation

[u

]S = P

[u

]T =

1 0 3−1 1 0

1 2 −1

9−1−8

= −15

−1017

∴(u

)S = (−15,−10,17)

c. Similarly, to find[u

]T using

[u

]S = (−6,7,2). That is to convert S coordinate into T coordinate. Hence

we use the translation matrix S to T , that is Q, given by the relation

[u

]T = Q

[u

]S = 1

10

1 −6 31 4 33 2 −1

−672

= 1

10

−4228−6

∴(u

)S =

(−21

5,

14

5,−3

5

)

Exercise 4.4

1. Find the coordinate of the following vectors relative to the basis S, given that,





a. v = (2,−1,3), S = v1, v2, v3

, v1 = (1,0,0) , v2 = (2,2,0) , v3 = (3,3,3)

b. p = 4−3x +x2, S = p1, p2, p3

, p1 = 1, p2 = x, p3 = x2

c. A =[

1 0−1 0

], S =

A1, A2, A3, A4

, A1 =

[ −1 10 0

], A2 =

[1 10 0

], A3 =

[0 01 0

], A4 =

[0 00 1

][Hint: c1 A1 + c2 A2 + c3 A3 + c4 A4 = 0]

2. Determine the coordinate vector of p = 4−2x +3x2 with respect to the standard basis of P2.

3. Consider the standard basis B for R3 and another basis C = (1,2,1) , (1,−1,1) , (1,0,−1):

a. Find the translation matrix P from C to B.

b. Find the translation matrix Q from B to C.

4. Consider the standard basis for P2 i.e. B = 1, x, x2 and another basis C =

2,−4x,5x2 −1

a. Find the translation matrices from C to B and B to C .

b. Determine the polynomial that has the coordinate vector(p

)C = (−4,3,11)

Answers

1. a. (3,−2,1) b. (4,−3,1) c.(−1

2,

1

2,−1,0

)2. (4,−2,3)

3. a. P = 1 1 1

2 −1 01 1 −1

b. Q = 1/6 1/3 1/6

1/3 −1/3 1/31/2 0 −1/2

4. a. PC→B = 2 0 −1

0 −4 00 0 5

, QB→c = 1/2 0 1/10

0 −1/4 00 0 1/5

b. p =−19−12x +55x2

E E E

4.14 Fundamental Spaces: Row Space, Column Space, Null Space

â Consider an m ×n matrix with real entries as,

A =

a 11

a 21

...

a m1

a 12

a 22

...

a m2

. . .

. . .

. . .

. . .

a 1n

a 2n

...

a mn

â The rows of above matrix are referred as row vectors of Rn and are denoted by r i , 1 É i É m

â The columns of above matrix are referred as column vectors of Rm and are denoted by c j , 1 É j É n

â Row space: The row space of the matrix A is defined as the span of row vectors of A and is denoted byrow(A). Hence,

row(A) = spanr 1,r 2,r 3.......r m

= k1r 1 +k2r 2 + ...+kmr m : k1,k2...km ∈R

â Column space: The column space of of the matrix A is defined as the span of column vectors of A andis denoted by col(A). Hence,

col(A) = spanc1,c2,c3.......cn

= k1c1 +k2c2 + ...+kncn : k1,k2...kn ∈R





â Null space: The solution space of the system of homogeneous system AX = 0 , is called the null spaceof A and is denoted by nul(A) . Hence,

nul(A) =

X ∈Rn : AX = 0

â Row space and Null space are subspace of of Rn , Column space is a subspace of Rm .

Theorem 4.2 (Basis for Row space and Column space)Let A be a given matrix and B be its equivalent row-echelon matrix. Then

i. The set of pivot rows of matrix B forms a basis for the row space of A.

ii. The set of columns of A, corresponding to the pivot columns of B forms a basis for the column spaceof A

4.15 Rank and Nullity

â Dimension of of row space of matrix A is called row rank of A.

â Dimension of of column space of matrix A is called column rank of A.

â The row rank and the column rank of a matrix are always same, and commonly it is known as rank ofmatrix A.

â Rank of A is given by number of pivots in the row echelon form, and is denoted by ρ(A).

â The dimension of the null space of A is called nullity of A and is denoted by µ (A) .

Alternatively, nullity is defined to be the number of non-pivot columns in the echelon form of matrixA or numbers of free variables in the solution of AX = 0..

â For any matrix A, rank(A) = rank(

AT ).

4.16 Rank-Nullity Theorem

Let A be an (m ×n) matrix then,

rank(A)+nullity(A) = Number of columns of A.

∴ ρ (A)+µ (A) = n

Note: Rank-Nullity theorem is also known as dimension theorem in context of ρ (A) = dim[row(A)] andµ (A) = dim[nul(A)]. Hence

dim[row(A)]+dim[nul(A)] = n = number of columns

Illustration 4.15 Find row(A) ,col(A) ,nul(A) , row(

AT ),col

(AT )

, nul(

AT ), given

A =

1 −2 1 1 2

−1 3 0 2 −20 1 1 3 41 2 3 13 5

Solution: Since

A =

1 −2 1 1 2

−1 3 0 2 −20 1 1 3 41 2 3 13 5

⇔ AT =

1 −1 0 1

−2 3 1 21 0 1 31 2 3 132 −2 4 5





1. row(A) =span of row vectors of A

= k1 (1,−2,1,1,2)+k2 (−1,3,0,2,−2)+k3 (0,1,1,3,4)+k4 (1,2,3,13,5) : k1,k2,k3,k4 ∈R

= (k −k2 +k4,−2k1 +3k2 +k3 +2k4,k1 +k3 +3k4,k1 +2k2 +3k3 +13k4,2k1 −2k2 +4k3 +5k4)

=col(

AT ) [∵ Rows of A are columns of AT ]

2. col(A) =span of column vectors of A

= k1 (1,−1,0,1)+k2 (−2,3,1,2)+k3 (1,0,1,3)+k4 (1,2,3,13)+k5 (2,−2,4,5) : k1, ...k5 ∈R

= (k1 −2k2 +k3 +k4 +2k5,−k1 +3k2 +2k4 −2k5,k2 +k3 +3k4 +4k5,k1 +2k2 +3k3 +13k4 +5k5)

=row(

AT ) [∵ Columnss of A are rows of AT ]

3. nul(A) =

X ∈R5 : AX = 0

.

Consider the augmented matrix for homogeneous system AX = 0:

x1 x2 x3 x4 x5

[A : Z ] =

1 −2 1 1 2 0

−1 3 0 2 −2 00 1 1 3 4 01 2 3 13 5 0

Reducing to row echelon form, we get

x1 x2 x3 x4 x5

[A : Z ] ∼

1 −2 1 1 2 00 1 1 3 0 00 0 −2 0 3 00 0 0 0 4 0

∴ ρ (A) = ρ (A : Z ) = 4 < 5 (number of unknowns)

∴ System has non trivial one parametric solution given by assigning parameter t to free variable x4,and is given by back substitution, as

x1 =−7t , x2 =−3t , x3 = 0, x4 = t , x5 = 0, t ∈R∴ X = (x1, x2, x3, x4, x5) = (−7t ,−3t ,0, t ,0) , t ∈R

∴ nul(A) = (−7t ,−3t ,0, t ,0) : t ∈R = span(−7,−3,0,1,0)

4. nul(

AT )= X ∈R4 : AT X = 0

.

Consider the augmented matrix for homogeneous system AT X = 0:

x1 x2 x3 x4

[AT : Z

]=

1 −1 0 1 0−2 3 1 2 0

1 0 1 3 01 2 3 13 02 −2 4 5 0

Reducing to row echelon form, we get

x1 x2 x3 x4

[AT : Z

]∼

1 −1 0 1 00 1 1 4 00 0 4 3 00 0 0 −2 00 0 0 0 0

∴ ρ (A) = ρ (A : Z ) = 4 (= number of unknowns)





∴ System has trivial solution, that is x1 = x2 = x3 = x4 = 0.Hence, nul

(AT )= (0,0,0,0) =

0

Illustration 4.16 Find the basis for the row space, column space and null space of the following matrix:[Summer-2016]

1 −3 4 −2 5 41 −6 9 −1 8 22 −6 9 −1 9 7

−1 3 −4 2 −5 −4

Solution: Let A =

1 −3 4 −2 5 41 −6 9 −1 8 22 −6 9 −1 9 7

−1 3 −4 2 −5 −4


A ∼

1 −3 4 −2 5 40 −3 5 1 3 −20 0 1 3 −1 −10 0 0 0 0 0

= B (4.10)

By Theorem (4.2),

1. Basis for row space of A is given by the set pivot rows of B . Hence basis for row space of A is

(1,−3,4,−2,5,4) , (0,−3,5,1,3,−2) , (0,0,1,3,−1,−1)

2. Basis for column space of A is given by the set of columns of A corresponding to pivot columns of B .Hence basis for row space of A is

(1,1,2,−1) , (−3,−6,−6,3) , (4,9,9,−4)

3. To find basis for null space of A, first we find null space of A.

From equation (4.10), the row echelon form of augmented matrix for system AX = 0 is

x1 x2 x3 x4 x5 x6

[A : Z ] ∼

1 −3 4 −2 5 40 −3 5 1 3 −20 0 1 3 −1 −10 0 0 0 0 0

0000

∴ ρ (A) = ρ (A : Z ) = 3 < 6 (number of unknowns)

∴ By back substitution, 3-parametric solution is given by

x1 =−s −5t , x2 =−14

3r + 8

3s + t , x3 =−3r + s + t , x4 = r, x5 = s, x6 = t , r, s, t ∈R

∴ X = (x1, x2, x3, x4, x5, x6) =(−s −5t ,−14

3r + 8

3s + t ,−3r + s + t ,r, s, t

)Hence,

nul(A) =(

−s −5t ,−14

3r + 8

3s + t ,−3r + s + t ,r, s, t

): r, s, t ∈R

â To find basis, rewrite nul(A) as a span of coefficient vectors of r, s, t , as follow:

nul(A) =

r

(0,−14

3,−3,1,0,0

)+ s

(−1,

8

3,1,0,1,0

)+ t (−5,1,1,0,0,1) : r, s, t ∈R

∴ nul(A) = span

(0,−14

3,−3,1,0,0

),

(−1,

8

3,1,0,1,0

), (−5,1,1,0,0,1)

Thus, basis for null space of A is

(0,−14

3,−3,1,0,0

),

(−1,

8

3,1,0,1,0

), (−5,1,1,0,0,1)

, because this

set is always linearly independent and it span nul(A) .





* Important:

From (4.10),

â Rank of A, ρ (A) = 3 = number of pivot columns.

â Nullity of A, that is dimension of null space of A, µ (A) = 3 = number of non-pivot columns.

â ρ (A)+µ (A) = 3+3 = 6 = total number of columns of A.

Hence, rank-nullity theorem is also verified.

Illustration 4.17 Find the basis for the vector space span(1,−1,2) , (0,5,−8) , (3,2,−2) , (8,2,0) .

Solution: Let W = span(1,−1,2) , (0,5,−8) , (3,2,−2) , (8,2,0) .If we consider a matrix A by putting the vectors in column, then W becomes column space of A. (Alter-

nately, we can also put vectors in rows, then W becomes row space of S)

A = 1 0 3 8

−1 5 2 21 −8 −2 0

⇒ W = col(A)

∴ Required basis for W , is the basis for column space of A.Reducing A in to row echelon form, we get

A ∼

1 0 3 80 5 5 100 0 3 8

= B

∴ Required basis is given by column vectors of A corresponding to non-pivot columns of B.∴ Basis for given vector space is (1,−1,1) , (0,5,−8) , (3,2,−1) .

Exercise 4.5

1. Find the basis for the row space, column space and null space of the following matrices and verify therank-nullity theorem:

a.

2 −4 1 2 −2 −3−1 2 0 0 1 −110 −4 −2 4 −2 4

b.

1 3 2 0 1

−1 −1 −1 1 00 4 2 4 31 3 2 −2 0

2. Find the basis for the vector space spanned by

[ −1 1−2 1

],

[2 −24 −2

],

[2 −13 1

],

[ −5 4−9 −1

][Hint: Consider matrix A by taking given matrices in column. Required basis is basis for col(A).]

3. Find basis for the space span1+x +x2 +x3,1+x2, 1+2x +2x2 +x3,1+x2 +2x3⊂P3 (x)

[Hint: Consider matrix A by taking coefficients in column. Required basis is basis for col(A).]

4. Let A be (3×4) matrix with ρ (A) = 3.

a. What is the dimension of

X : AX = 0

?

b. Is AX = b consistent for all b ?

c. If AX = b is consistent, how many solution does it have ?

5. Prove that the null space of m ×n matrix is a subspace of Rn .

6. Find rank and nullity of the matrix A = 2 0 −1

4 0 −20 0 0

and verify dimension theorem. [Summer-2015]





Answers

1. a. Row space basis: (2,−4,1,2,−2,−3) , (0,16,−7,−6,8,19) ,

(0,0,

1

2,1,0,−5

2

),

Column space basis: (2.−1,10) , (−4,2,−4) , (1,0,−2),

Null space basis:

(−1,−1

2,−2,1,0,0

),

(0,−1

2,0,0,1,0

), (1,1,5,0,0,1)

, Rank 3, Nullity 3.

b. Row space basis: (1,3,2,0,1) , (0,2,1,1,1) , (0,0,0,2,1),Column space basis: (1,−1,1,0) , (1,−1,4,3) , (0,1,4,−2),

Null space basis:

(−1

2,−1

2,1,0

),

(−1

4,−1

4,−1

2,1

), Rank 3, Nullity 2.

2.[ −1 1

−2 1

],

[2 −13 1

],

[ −5 4−9 −1

]3.

1+x +x2 +x3,1+x2, 1+2x +2x2 +x3,1+x2 +2x3

4. a. 1 b. No c. Infinite 6. Rank 1, Nullity 2

E E E

Powered by





Contact: (+91) 9427 80 9779





Chapter 5Linear Transformation (Linear Mapping)

5.1 Linear transformation

Let V and W are real vector spaces then a mapping or a function or a transformation defined from V to Wthat is T : V →W is said to be linear transformation if it will satisfies the following two conditions.

i. ∀u, v ∈V ; T(u + v

)= T(u

)+T(v)

ii. ∀u ∈V ,α ∈R; T(αu

)=αT(u

)* Important:

1. T : V →W preserves two basic operations of the vector space namely vector addition and scalar mul-tiplication.

2. For α= 0, T(0v

)= 0w . Hence linear transformation maps zero vector V to the zero vector of W .

3. For V = Rn and W = Rm , linear transformation T : Rn → Rm is also known as Euclidian linear trans-formation. If m = n , T :Rn →Rn is called linear operator on Rn

5.2 Particular Transformations

1. Zero transformation: A linear transformation T : V →W is c zero transformation if, T(v)= 0, ∀∈V.

2. Identity transformation: An operator T : V →V is said to be an identity operator if, T(v)= v , ∀∈V.

Illustration 5.1 Check whether following mapping are linearly transformation or not.

a. T :R2 →R2, T(x, y

)= (x +2y,3x − y

)[summer-2016]

b. T :R3 →R2, T(x, y, z

)= (|x| , y + z)

c. T : M22 →R2, T

[a bc d

]=

(∣∣∣∣ a bc d

∣∣∣∣ ,0

)Solution:a. Let u = (

x1, y1)

, v = (x2, y2

) ∈R2, α ∈R

∴ T(u

)= T(x1, y1

)= (x1 +2y1,3x1 − y1

),T

(v)= T

(x2, y2

)= (x2 +2y2,3x2 − y2

)i. T

(u + v

)=T(x1 +x2, y1 + y2

)= [

(x1 +x2)+2(y1 + y2

),3(x1 +x2)− (

y1 + y2)]

= (x1 +x2 +2y1 +2y2,3x1 +3x2 − y1 − y2

)= (

x1 +2y1,3x1 − y1)+ (

x2 +2y2,3x2 − y2)

∴ T(u + v

)= T(u

)+T(v)

51




ii. T(αu

)=T(αx1,αy1

)= [

(αx1)+2(αy1

),3(αx1)− (

αy1)]

= (αx1 +2αy1,3αx1 −αy2

)=α(

x1 +2y1,3x1 − y1)

∴ T(αu

)=αT(u

)∴ T preserves vector addition and scalar multiplication. Hence T is a linear transformation.

b. u = (x1, y1, z1

), v = (

x2, y2, z2) ∈R3, α ∈R

i. T(u + v

)=T(x1 +x2, y1 + y2, z1 + zz

)= (|x1 +x2| , y1 + y2 + z1 + zz

) [By given definition

]6= (|x1| , y1 + z1

)+ (|x2| , y2 + zz)

[∵ |x1 +x2| 6= |x1|+ |x2|]∴ T

(u + v

) 6= T(u

)+T(v)

∴ T does not preserve vector addition. Hence T is not linear transformation.

c. Given that T

[a bc d

]=

(∣∣∣∣ a bc d

∣∣∣∣ ,0

)= (ad −bc,0)

∴ u =[

a1 b1

c1 d1

]; v =

[a2 b2

c2 d2

]∈ M22 ⇒ T

(u

)= (a1d1 −b1c1,0) ,T(v)= (a2d2 −b2c2,0)

i. T(u + v

)=T

[a1 +a2 b1 +b2

c1 + c2 d1 +d2

]=

(∣∣∣∣ a1 +a2 b1 +b2

c1 + c2 d1 +d2

∣∣∣∣ ,0

)= ((a1 +a2) (d1 +d2)− (b1 +b2) (c1 + c2) ,0)

= (a1d1 +a2d1 +a2d2 −b1c1 −b1c2 −b2c1 −b2c2,0)

6= (a1d1 −b1c1,0)+ (a2d2 −b2c2,0)

T(u + v

) 6=T(u

)+T(u

)∴ T does not preserve vector addition. Hence T is not linear transformation.

* Important:

â For linear transformation, formula of T must be linear, that is of the form ax +by + cz, other wisemapping is not linear transformation.

â In formula of T , if there is some non linear term like product, power, modulus, constant, any nonlinear function etc. then T be never linear transformation.

â For example, T : R3 → R2 defined by T(x, y, z

) = (x −2y,4x + y +3z

)is linearly transformation but

T(x, y, z

) = (x y,4x + y +3z

)or T

(x, y, z

) = (x − y, z2 +1

)or T

(x, y, z

) = (x + y −2z, tan x

)etc. are not

linear transformations. (verify !)

Illustration 5.2 Determine the linearly transformation T :R2 →R3 such that T (1,0) = (1,2,3) and T (1,1) =(0,1,0). Also find T (2,3) . [Summer-2016]

Solution: To find general formula for T : R2 → R3, first we represent an arbitrary vector(x, y

) ∈ R2 as alinear combination of (1,0) and (1,1) . for this consider(

x, y)= c1 (1,0)+ c2 (1,1) ⇒ c1 + c2 = x, c2 = y ∴ c1 = x − y, c2 = y





Hence,(x, y

)=(x − y

)(1,0)+ y (1,1)

⇒ T(x, y

)=T[(

x − y)

(1,0)+ y (1,1)] [

Applying T on both sides]

=(x − y

)T (1,0)+ yT (1,1) [∵ T is linear transformation]

=(x − y

)(1,2,3)+ y (0,1,0) [∵ Given T (1,0) = (1,2,3) ,T (1,1) = (0,1,0)]

=(x − y,2x −2y,3x −3y

)+ (0, y,0

)∴ T

(x, y

)=(x − y,2x − y,3x −3y

)Required formula.

Now put(x, y

)= (2,3) ∴ T (2,3) = (−1,1,−3)

5.3 Matrix Linear Transformation

â Let A be an m×n matrix then its induced linear transformation TA :Rn →Rm is defined as TA(x)= Ax.

â Further, if T :Rn →Rm be a linear transformation then there exit an m ×n matrix A such that T = TA ,that is T

(x)= Ax.

â Matrix A is called matrix of T or standard matrix of T and is sometimes denoted by A = [T ].

Illustration 5.3

a. Find matrix of the linearly transformation T :R4 →R3 defined by,

T(w, x, y, z

)= (w −2x − y +2z,−2w +4x +3y − z,−w +2x + y − z

).

b. Find linearly transformation induced by the matrix

−2 1 43 5 76 0 −1

.

Solution:a. Given T : R4 → R3, so induced (standard) matrix A is of the order (3×4), which can be constructed byfollowing method:

â There are four unknowns w, x, y, z in definition of T , so A has four columns.

â Formula of T gives three linear equations, so A has three rows.

â Induced matrix A can be constructed by putting coefficients of w, x, y, z in rows respectively, as bellow:

w x y z

A = 1 −2 −1 2

−2 4 3 −1−1 2 1 −1

= [T ]

b. Let A = −2 1 4

3 5 76 0 −1

.

Since A is (3×3) matrix, the induced transformation is TA : R3 → R3 and is defined by TA

(X

)= AX , X ∈ R3.

Hence,

∀X = x

yz

∈R3, TA

xyz

= −2 1 4

3 5 76 0 −1

xyz

= −2x + y +4z

3x +5y +7z6x − z

∴ TA

(x, y, z

)= (−2x + y +4z,3x +5y +7z,6x − z)

,(x, y, z

) ∈R3

Exercise 5.1

1. Determine whether the following mappings are linearly transformation or not?





a. T :R3 →R2, T(x, y, z

)= (y, z

)b. T :R2 →R2, T

(x, y

)= (x +2, y2)

c. T :R2 →R2, T(x, y

)= (x y, x

)d. T :R2 →R3, T

(x, y

)= (x +3,2y, x + y

)e. T : M22 →R2, T

[a bc d

]= (ad +1,b + c) f. T : M22 →R2, T

[a bc d

]= (2a −b −d ,0)

2. Determine the linearly transformation T :R2 →R3 such that T (−1,1) = (1,1,2) and T (3,−1) = (−2,0,1).

3. If S = e1, e2, e3 be the standard basis for R3 and T : R3 → R3 be linearly transformation such thatT (e3) = 2e1 +3e2 +5e3 , T (e2 + e3) = e1 and T (e1 + e2 + e3) = e2 − e3 then find T (e1 +2e2 +3e3).

[Hint: T (e1 +2e2 +3e3) = T (e1)+T (e2 + e3)+T (e1 + e2 + e3)]

4. S = u1,u2,u3

where u1 = (1,1,1) ,u2 = (1,1,0) ,u3 = (1,0,0) be the basis for R3 Let T : R3 → R2 be the

LT. Assume that T(u1

)= (1,2) ,T(u2

)= (3,4) and T(u3

)= (5,6) . Find the formula for T (x1, x2, x3).

5. Find standard matrix for the following linearly transformation :

a. w1 =2x1 −3x2 +x3

w2 =3x1 +5x2 −x3

b. w1 =x1

w2 =x1 +x2

w3 =x1 +x2 +x3

w4 =x1 +x2 +x3 +x4

[Hint: Take W = T(

X)]

Answers

1. a. c. f. are LT. 2. T(x, y

)= 1

2

(−x + y, x +3y,3x +7y)

3. (3,4,4)

4. T(x, y, z

)= (5x −2y −2z,6x −2y −2z

)5. a.

[2 −3 13 5 −1

]b.

1 0 0 01 1 0 01 1 1 01 1 1 1

E E E

5.4 Composition Linear Transformations

â If T : V → W and S : W →U are linear transformations then the composition S T : V →U is also alinear transformation.

â Composition of two linear transformations is also denoted by ST.

â In particular, if TA : Rn → Rk and TB : Rk → Rm are induced by matrices A and B respectively then(TB TA) :Rn →Rm is composite linearly transformation and is defined as (TB TA) (X ) = (B A) X . Alsothe matrix transformation is given by [TB TA] = [TB ] [TB ] = B A

Illustration 5.4 Show that T :R2 →R2 and S :R2 →R3 defines as T (a,b) = (a +b,b) and S (a,b) = (2a,b, a +2b)are linear. Also find formula for composite transformation S T .

Solution: Given T (a,b) = (a +b,b) and S (a,b) = (2a,b, a +2b). Clearly T and S are linear. (verify !)Consider induced matrices of T and S, given by

[T ] =[

1 10 1

]= A, [S] =

2 00 11 2

= B





The induced matrix for composite transformation S T is defined by

[S T ] = [S] [T ] = B A

= 2 0

0 11 2

[1 10 1

]=

2 20 11 3

∴ [S T ] = 2 2

0 11 3

=C(Say

)

∴ formula for S T :R2 →R3 is

S T(

X)=C X , X ∈R2

= 2 2

0 11 3

[ab

]

= 2a +2b

ba +3b

∴ S T (a,b) = (2a +2b,b, a +3b) Required formula.

5.5 Onto (Surjective) Linear Transformations

A linear transformation T : V →W is said to be onto (surjective), if for all w ∈W there exist an element v ∈Vsuch that T

(v)= w .

5.6 One-one (Injective) Linear Transformations

A linear transformation T : V →W is said to be one-one (injective) , if ∀u, v ∈V ; T(u

)= T(v) ⇒ u = v

or ∀u, v ∈V ; u 6= v ⇒ T(u

) 6= T(v)

* Important:

Let T :Rn →Rm be linearly transformation and A be its induced matrix transformation then,

1. T is onto if AX = b has solution for all b, that is all rows of row echelon form of A are pivots.

2. T is one-one if AX = 0 has unique solution, that is all columns of row echelon form of A are pivots.

3. A linearly transformation which is one-one and onto is known as one-onto-one or Bijective.

Illustration 5.5 Check whether T : R3 → R3 defined by T(x, y, z

) = (x +3y, y,2x + z

)is linear. Is it one to

one and onto ? [Summer-2017]

Solution: Given T(x, y, z

)= (x +3y, y,2x + z

).

Let u = (x1, y1, z1

), v = (

x2, y2, z2) ∈R3, α ∈R

i. T(u + v

)=T(x1 +x2, y1 + y2, z1 + z2

)=(

(x1 +x2)+3(y1 + y2

),(y1 + y2

),2(x1 +x2)+ (z1 + z2)

) [By given definition

]=(

x1 +x2 +3y1 +3y2, y1 + y2,2x1 +2x2 + z1 + z2)

=(x1 +3y1, y1,2x1 + z1

)+ (x2 +3y2, y2,2x2 + z2

)∴ T

(u + v

)=T(u

)+T(v)





ii. T(αu + v

)=T(αx1,αy1,αz1

)=(α

(x1 +3y1

),α

(y1

),α (2x1 + z1)

)=(αx1 +3αy1,αy1,2αx1 +αz1

)=α(

x1 +3y1, y1,2x1 + z1)

∴ T(αu

)=αT(u

)∴ T preserves vector addition and scalar multiplication. Hence T is a linear transformation.Now for one-one and onto, we reduce the standard matrix of T to row echelon form:

A = 1 3 0

0 1 01 0 1

→ R3 −R1

∼ 1 3 0

0 1 00 −3 1

→ R3 +3R2

∼

1 3 00 1 00 0 1

= B

Observe that, in echelon form B all columns and rows are pivot. Hence given linearly transformation isone-one and onto.Note: Here T is both one-one and onto, hence it is bijective.

5.7 Range (Image) and Kernel

Let T : V →W be the linear transformation, then

1. The range of T is defined as Range(T ) = R (T ) = T

(v)

: ∀v ∈V= Im(T )

2. The kernel of T is defined as ker(T ) = v ∈V : T

(v)= 0w

* Important:

In particular, if T :Rn →Rm and A be its induced matrix, that is T(

X)= AX , X ∈Rn , then

1. Range of T : R (T ) =

AX : X ∈Rn= col(A) = col(T ) [Known column space of T ]

2. Kernel of T : ker(T ) =

X ∈Rn : AX = 0= nul(A) = nul(T ) [Known null space of T ]

Remark:

1. Range of T : V →W is subspace of W and Kernel is subspace of V

2. For zero linear transformation, that is T(v)= 0w , ∀v ∈V ,

R (T ) = 0w

and ker(T ) =V

3. For identity linear operator, that is T(v)= v , ∀∈V ,

R (T ) =V and ker(T ) = 0v

Theorem 5.1 (Rank-Nullity theorem or Dimension theorem for linear transformation)

Let T :Rn →Rm be ıand A be its induced matrix, that is T(

X)= AX , X ∈Rn , then rank and nullity of T are

defined as rank and nullity of A respectively. Hence,

rank(A)+nullity (A) = n = number of columns ⇔ rank(T )+nullity (T ) = dimRn





Rank-Nullity theorem is also known as dimension theorem in context of rank(T ) = dimR (T ) and nullity (T ) =dim[nul(T )]. Hence

dim[R (T )]+dim[nul(T )] = dimRn

Illustration 5.6 Show that the mapping T : R2 →R3 defined by T(x, y

)= (x + y, x − y, y

)is linear transfor-

mation. Find range, null space (kernel), rank and nullity of T .

Solution: Given T(x, y

)= (x + y, x − y, y

). Clearly, T is linear transformation. (Verify !)

Induced matrix for T is (3×2), given by,

A = 1 1

1 −10 1


A ∼

1 10 −20 1

1. Range: By definition,

R (T ) =

T(

X)

: X ∈R2= (

x + y, x − y, y)

: x, y ∈R2= col(T ) = col(A)

2. Null space (Kernel): By definition, ker(T ) =

X ∈R2 : T(

X)= 0

=

X ∈R2 : AX = 0

, that is kernel of

T is the solution space of AX = 0. From row echelon form of A, system has trivial solution ρ (A) = 2 =number of unknowns. Therefore x = 0, y = 0. Thus,

ker(T ) = 0= col(T ) = nul(A)

3. Rank: Rank of T = Rank of A = 2. (Dimension of null space of A, that is number of non pivot columns)

4. Nullity: Nullity of T = Nullity of A = 0. (Dimension of column space of A, that is number of pivotcolumns)

Illustration 5.7 State the Dimension theorem for linear transformation and find the rank and nullity of

TA , where TA :R6 →R4 be multiplication by A =

−1 2 0 4 5 −3

3 −7 2 0 1 42 −5 2 4 6 14 −9 2 −4 −4 7

[Winter-2017]

Solution: Statement og dimension theorem for linearly transformation is given by Theorem 5.1.Now rank and nullity of T are defined to be the rank and nullity of A. Reducing A to row echelon form,

we get

A =

−1 2 0 4 5 −3

3 −7 2 0 1 42 −5 2 4 6 14 −9 2 −4 −4 7

→ R2 +3R1; R3 +2R1; R4 +4R1

∼

−1 2 0 4 5 −3

0 −1 2 12 16 −50 −1 2 12 16 50 −1 2 12 16 5

→ R3 −R2; R4 −R2

∼

−1 2 0 4 5 −3

0 −1 2 12 16 −50 0 0 0 0 00 0 0 0 0 0





â Rank of T = Rank of A = 2 (Dimension of Column space T )

â Nullity of T = Nullity of A = Number of non pivot columns = 4 (Dimension of null space of T )

â Observe that, dim[col(T )]+dim[nul(T )] = 2+4 = 6 = dimR6.

Hence dimension theorem for linearly transformation is verified.

Illustration 5.8 Let T :R3 →R3 be linearly transformation defined as T(x, y, z

)= (x +2y − z, y + z, x + y −2z

).

Find a basis and dimension of the range and kernel of T .

Solution: Induced matrix for given linearly transformation is A = 1 2 −1

0 1 11 1 −2

.

Basis for Range (Image) and Kernel of T , are the basis for column space and null space of A respectively.Reducing A to row echelon form:

A = 1 2 −1

0 1 11 1 −2

→ R3 −R1

∼ 1 2 −1

0 1 10 −1 −1

→ R3 +R2

∼

1 2 −10 1 10 0 0

= B

1. Basis for Range of T is the set of columns of A corresponding to pivot columns of B , and is given by(1,0,1) , (2,1,1) . Also dimension of range of T is 2.

2. Basis for Kernel (null space) of T is the basis for null space of A, that is solution space of AX = 0, X ∈R3.From B , system AX = 0 has one parametric non trivial solution given by X = (3t .− t , t ) , t ∈R.

Hence basis for null space of T is (3.−1,1). Also dimension of kernel of T is 1.

Illustration 5.9 Let T :R3 →R3 be the LT defines as T(x, y, z

)= (x + y − z, x −2y + z,−2x −2y +2z)

a. Which of the vectors from the set (1,2,3) , (1,2,1) , (−1,1,2) belongs to ker(T ) ?

b. Which of the vectors from the set (1,2,−2) , (3,5,2) , (−2,3,4) belongs to R (T ) ?

Exercise 5.2

1. Determine whether the given LT be one-onto-one (bijective) or not ?

a. T :R4 →R3; T (a,b,c,d) = (a −2b − c +2d ,−2a +4b +3c −d ,−a +2b + c −d)

b. T :R3 →R4; T (a,b,c) = (a +3b +2c,−a −b − c,4b +2c, a +3b +2c)

c. T :R4 →R4; T (a,b,c,d) = (a +2b − c +2d ,2a +b +3c +2d , a −b +2c +2d ,2b +d)

d. T :R2 →R3; T (a,b) = (a +2b,2a +3b,3a +4b)

2. Let T :R4 →R3 be the linearly transformation defined as

T(x, y, z, w

)= (x − y + z +w, x +2z −w, x + y +3z −3w

)Find a basis and dimension of range and kernel of T.

3. Find the corresponding transformation and indicate the source (image) and the target (co-domain)Euclidean spaces for the matrices:

a.[

1 2]

b.

2 0 2 11 1 −1 00 1 −2 1





Answers

1. a. onto, not one-one b. Not one-one, Not onto c. one-one and onto d. one-one, not onto

2. Basis for Range: (1,1,1) , (−1,0,1) ,dim = 2; Basis for kernel: (−2,−1,1,0) , (1,2,0,1) ,dim = 2.

3. a. T :R2 →R, T(x, y

)= x +2y, Source: R2, Target: R

b. T :R4 →R3, T(x, y, z, w

)= (2x +2z +w, x + y − z, y −2z +w

), Source: R4, Target: R3

E E E

5.8 Inverse Linear Transformation (Isomorphism)

â A linearly transformation T : Rn → Rn is said to be invertible if there exist a linearly transforma-tion T −1 :Rn →Rn such that T T −1 = T −1 T = I

where I :Rn →Rn is an identity linear operator.

â Here T and T −1 are called inverse linearly transformation of each other.

â In this case, T(

X)= Y ⇔ Y = T −1

(X

)â A linear transformation is invertible if and only if it is bijective (one-one and onto).

â If A is an induced matrix of T , then T is invertible if and only if matrix A is invertible.

In this case formula for T −1 is given by

T −1(

X)= A−1X ⇔ [

T −1]= A−1

â An invertible (Bijective) transformation is called Isomorphism

â Two vector spaces V and W are said to be isomorphic if there exist an isomorphisam from v to W .

Illustration 5.10 Check whether T :R3 →R3 defined by T (x1, x2, x3) = (3x1+x3,−2x1+x2,−x1+2x2+4x3)is invertible (Isomorphism) or not. If invertible then find formula for T −1.

Solution: Given T (x1, x2, x3) = (3x1 +x3,−2x1 +x2,−x1 +2x2 +4x3). The matrix for T is given by

A = 3 0 1

−2 1 0−1 2 4

We know that T is invertible if and only if A−1 exist, that is A should be non singular. Since

det(A) =∣∣∣∣∣∣

3 0 1−2 1 0−1 2 4

∣∣∣∣∣∣= 3(4−0)−0+1(−4+1) = 9 6= 0

∴ A is non singular, that is A−1 exist. Hence T is invertible (Isomorphism) and formula for T −1 is

T −1(

X)= A−1X , X ∈R3. (5.1)

Now,

A−1 = 1

|A| adj(A) = 1

9

4 2 −18 13 −2

−3 −6 3





∴ From (5.1),

T −1(

X)= 1

9

4 2 −18 13 −2−3 −6 3

x1

x2

x3

= 1

9

4x1 +2x2 −x3

8x1 +13x2 −2x3

−3x1 −6x2 +3x3

∴ T −1 (x1, x2, x3) =

(4x1 +2x2 −x3

9,

8x1 +13x2 −2x3

9,−3x1 −6x2 +3x3

9

)

Exercise 5.3In each of the following case find T −1, if exist ?

1. T :R2 →R2, T(x, y

)= (2y,3x − y) 2. T :R3 →R3, T(x, y, z

)= (2y + z, x −4y,3x

)3. T :R3 →R3, T (x1, x2, x3) = (x1 −x2, x2 −x1, x1 −x3)

Answers

1. T −1 (x, y

)= (x +2y

6,

x

2

)2. T −1 (

x, y, z)= ( z

3,− y

4+ z

12, x + y

2− z

6

)3. T −1 does not exist.

E E E

Powered by





Contact: (+91) 9427 80 9779





Chapter 6Eigenvalues and Eigenvectors

6.1 Definition

â Let A be the square matrix of order n. A non-zero vector X ∈Rn is said to be an eigenvector of A if therea scalar λ (real or complex) such thatAX =λX .

â The scalar λ is said to be eigenvalue or characteristic value of A and the vector X is said to be eigen-vector or characteristic vector of A corresponding to the eigenvalue λ.

6.2 Method of Finding Eigenvalue and Eigenvector

Suppose λ be the eigenvalue corresponding the non-zero eigenvector←−X , then

AX =λX(

X 6= 0)

⇒ AX =λI X , where I is an identity matrix.

⇒ AX −λI X = 0

⇒ (A−λI ) X = 0(

X 6= 0)

(6.1)

This is homogeneous system of equations which has non-trivial solution. Hence

det(A−λI ) = | A−λI | = 0 (6.2)

â Equation (6.2) is called characteristic equation or characteristic polynomial of materix A.

â On solving equation (6.2), we get n eigenvalues as λ1,λ2,λ3.....λn .

â Solving equation (6.1), for each value of λ we get corresponding eigenvector.

6.3 Properties of Eigenvalues

1. The set of all eigenvalues of A is called the spectrum of A.

2. Trace of A = Sum of all eigenvalues of A.

3. If λ1,λ2,λ3.....λn are the eigenvalues of A then λ1 ×λ2 ×λ3 × .....×λn = | A | = det(A) .

4. The eigenvalues of upper or lower triangular matrix, hence the diagonal matrix are the elements ofits main diagonal.

5. If λ is an eigenvalues of a non singular matrix A then1

λis an eigenvalue of A−1.

6. If λ is an eigenvalues of A then kλ is an eigenvalue of k A.

61




7. If λ is an eigenvalues of A then λm is an eigenvalue of Am where m ∈N.

8. Spectral shift: If λ is an eigenvalue of A, then λ−k is eigenvalue of matrix (A−kI ) .

9. Matrices A and AT have same eigenvalues.

10. A matrix is a non singular if and only if λi 6= 0, ∀i = 1,2,3.....n. OR A square matrix A is invertible ifand only if λ= 0 is not an eigenvalue of A.

11. If λ is an eigenvalues of a non singular matrix A then|A|λ

is an eigenvalue of adj A.

12. The eigenvalues of symmetric matrix are real.

6.4 Properties of Eigenvectors

1. The eigenvector corresponding to the eigenvalue is not unique. That is if X is an eigenvector corre-sponding to the eigenvalue λ so is k X , for the scalark 6= 0 .

2. If λ1,λ2,λ3.....λn are the distinct eigenvalues of an (n ×n) matrix then the corresponding eigenvectorsX 1, X 2, X 3.......X n are linearly independent.

3. When two or more eigenvalues are equal it may or may not be possible to get linearly indepen-dent eigenvectoreigenvectors corresponding repeated eigenvalues.

4. eigenvalue may be zero but eigenvector can not be zero.

5. All eigenvectors of a symmetric matrix are always linearly independent and orthogonal.

6. Eigen space: Let λ be an eigenvalue of the matrix A then the set Eλ =

X : AX =λX

is called the eigen

space of λ.

In other words, the solution space of the system (A−λI ) X = 0 or null space of the matrix transfor-mation (A−λI ) is called the eigen space.

* Important:

1. If A be the square matrix of order (2×2) then its characteristic equation is given by,

λ2 −S1λ+|A| = 0,

whereS1 = trace(A) = Sum of diagonal elements = a11 +a22, |A| = det(A)

2. If A be the square matrix of order (2×2) then its characteristic equation is given by,

λ3 −S1λ2 +S2λ−|A| = 0,

whereS1 = trace(A) = Sum of diagonal elements = a11 +a22 +a33,

S2 = Sum of minors of diagonal elements = M11 +M22 +M33

|A| = det(A)

3. Cramer’s Rule:

a1x +b1 y + c1z = 0

a2x +b2 y + c2z = 0

⇒ x∣∣∣∣ b1 c1

b2 c2

∣∣∣∣=− y∣∣∣∣ a1 c1

a2 c2

∣∣∣∣= z∣∣∣∣ a1 b1

a2 b2

∣∣∣∣= t

(Say

), t ∈R





Illustration 6.1 Find the eigenvalues and eigenvectors of the matrix

[14 −105 −1

].

Solution: Suppose A =[

14 −105 −1

].

If λ be the eigenvalue of A corresponding to non zero eigenvector←−X , then

AX =λX ,(

X 6= 0 ∈R2)

⇒ (A−λI ) X = 0 (6.3)

The characteristic polynomial of system (6.3) is

det(A−λI ) = 0

⇒ λ2 −S1λ+|A| = 0,

where S1 = trace(A) = a11 +a22 = 14−1 = 13,

|A| = det(A) =−14+50 = 36

∴ λ2 −13λ+36 = 0 ⇒ (λ−4)(λ−9) = 0

∴ λ= 4,9 =λ1,λ2 (Say)

∴ eigenvalues of A are 4,9.

Now, eigenvectors of A are given by equation (6.3), as

(A−λI ) X = 0

∴([

14 −105 −1

]−λ

[1 00 1

])[xy

]=

[00

]∴

([14 −105 −1

]−

[λ 00 λ

])[xy

]=

[00

]∴

[14−λ −10

5 −1−λ][

xy

]=

[00

](6.4)

λ === λ1 === 4 : Substituting λ = 4 in equation (6.4), we get

[10 −105 −5

][xy

]=

[00

]. This yields one

equation10x −10y = 0 ⇒ x = y = t

(say

), t ∈R

The corresponding eigenvector is obtained by taking any non zero value of t . Without loss of generality (for

simplicity), we take t = 1. Hence the eigenvector for λ1 = 4 is X 1 =[

11

].

λ===λ2 === 9 : Substituting λ= 9 in equation (6.4), we get[5 −105 −10

][xy

]=

[00

]⇒ 5x −10y = 0 ∴

x

2= y = t

(say

), t ∈R

For t = 1, eigenvectoreigenvector corresponding to λ2 = 9 is X 2 =[

21

]. Thus,

λ1 = 4 → X 1 =[

11

], λ2 = 9 → X 2 =

[21

]

Illustration 6.2 Find the eigenvalues and basis for eigen space for the matrix A = 3 −1 0

−1 2 −10 −1 3

.

[Summer-2016]





Solution: Ifλ be the eigenvalue of A corresponding to eigenvector X ∈R3, then the characteristic equationis,

det(A−λI ) = 0 ⇒ λ3 −S1λ2 +S2λ−|A| = 0 (6.5)

where S1 =trace(A) = Sum of diagonal elements = a11 +a22 +a33

= 3+2+3 = 8,

S2 =Sum of minors diagonal elements = M11 +M22 +M33

=∣∣∣∣ 2 −1−1 3

∣∣∣∣+ ∣∣∣∣ 3 00 3

∣∣∣∣+ ∣∣∣∣ 3 −1−1 2

∣∣∣∣= 5+9+5 = 19,

|A| =∣∣∣∣∣∣

3 −1 0−1 2 −10 −1 3

∣∣∣∣∣∣= 3(6−1)+1(−3−0)+0 = 15−3 = 12.

Substituting in (6.5), we get characteristic equation

λ3 −8λ2 +19λ−12 = 0 (6.6)

Method to find roots of characteristic equation: Equation (6.6) has cubic polynomial, so it may hasthree real roots. To find these roots one of the following method:

â Method 1: Put different values of λ say 0,1,−1,2,−2,3,−3.... until equation is satisfied (that is LHSbecomes 0).

Observe that forλ= 1, equation (6.6) is satisfied. Hence one root isλ= 1. That is one factor is (λ−1).

â To find other roots adjust second factor of characteristic polynomial as

λ3 −8λ2 +19λ−12 =λ2 (λ−1)−7λ (λ−1)+12(λ−1)

= (λ−1)(λ2 −7λ+12

)= (λ−1)(λ−3)(λ−4)

â Method 2: Since one value of λ is 1 (i.e. one factor is (λ−1)), second factor can be obtained usingSynthetic Division (click here) as follow:

1 1 −8 19 −120 1 −7 121 −7 12 0

∴ Second factor is λ2 −7λ+12. Hence, λ3 −8λ2 +19λ−12 = (λ−1)(λ2 −7λ+12

).

From (6.6),

λ3 −8λ2 +19λ−12 = 0

∴ (λ−1)(λ2 −7λ+12

)= 0

∴ (λ−1)(λ−3)(λ−4) = 0

∴ λ= 1,3,4 =λ1,λ2,λ3(Say

)∴ eigenvalues of A are 1,3,4.

Now the eigenvector X 6= 0 ∈R3 is given by homogeneous system,

(A−λI ) X = 0,(

X 6= 0 ∈R3)

⇒ 3 −1 0

−1 2 −10 −1 3

−λ 1 0 0

0 1 00 0 1

xyz

= 0

00


https://www.youtube.com/watch?v=x4xuFDg4Esc&feature=youtu.be




∴

3 −1 0−1 2 −10 −1 3

− λ 0 0

0 λ 00 0 λ

xyz

= 0

00

∴

3−λ −1 0−1 2−λ −10 −1 3−λ

xyz

= 0

00

(6.7)

λ===λ1 === 1 : From (6.7),

2 −1 0−1 1 −1

0 −1 2

xyz

= 0

00

.

Since the system has non trivial solution, this yields exactly two different equations. Solution of thissystem is obtained by considering any two different equations out of three rows. Considering 1st and 3rdrow, we get

2x − y = 0, −y +2z = 0 ⇒ 2x = y = 2z ∴ x = y

2= z = t , t ∈R

For t = 1, eigenvector corresponding to eigenvalue λ= 1 is X 1 = 1

21

.

Eigen space: By definition, eigen space of λ1 = 1 is the solution space of the system (A−λ1I ) X = 0. That isthe set of all one parametric solutions of the system (A−λ1I ) X = 0. Hence

Eλ1 =

X ∈R3 : (A−λ1I ) X = 0= (t ,2t , t ) : t ∈R

∴ Eλ1 = span(1,2,1)

Since the singleton set (1,2,1) is linearly independent (because vector in not zero) and it spans the eigenspace Eλ1 , hence it is basis for eigen space Eλ1 .∴ Basis for eigen space Eλ1 = (1,2,1)â It is worth to note that the basis for eigen space is the set of corresponding eigenvector.

λ===λ2 === 3 : From (6.7),

0 −1 0−1 1 −10 −1 0

xyz

= 0

00

∴ From 1st and 2nd equation,

−y = 0, −x + y − z = 0 ⇒ y = 0, x =−z = t , t ∈R

For t = 1, eigenvector for λ2 = 3 is X 2 = 1

0−1

â Eigen space Eλ2 = (t ,0,−t ) : t ∈R and basis for eigen space is (1,0,−1) .

λ===λ3 === 4 : From (6.7), ∴

−1 −1 0−1 −2 −10 −1 −1

xyz

= 0

00

∴ From 1st and 3rd equations,

−x − y = 0, −y − z = 0 ⇒ −x = y =−z = t , t ∈R

For t = 1, eigenvector for λ3 = 4 is X 3 = −1

1−1

â Eigen space Eλ2 = (−t , t ,−t ) : t ∈R and basis for eigen space is (−1,1,−1) .

Important deductions:

Using properties of eigenvalue [See section 6.3], we have following important deductions:





1. Spectrum of A : (1,2,1) , (1,0,−1) , (−1,1,−1) .

2. λ1 +λ2 +λ3 = 1+3+4 = 7 = Trace of A.

3. λ1 ×λ2 ×λ3 = 1×3×4 = 12 = det(A) .

4. Here λ= 1,3,4 are eigenvalues of A, so

i. λ = 0 is not an eigenvalue of A. Hence A is non singular matrix and eigenvalues of A−1 are

λ−1 = 1,1

3,

1

4.

ii. eigenvalues of 3A are 3λ= 3,9,12, of −2A are −2λ=−2,−6,−8 and so on.

iii. eigenvalues of A2 are λ2 = 1,9,16, of A3 are λ3 = 1,27,64 and so on.

iv. eigenvalues of A−3I are λ−3 =−2,0,1, of A+4I are λ+4 =−5,7,8 and so on.

v. eigenvalues of AT are 1,3,4.

vi. eigenvalues of adj(A) are|A|λ

= 12

1,

12

3,

12

4= 12,4,3.

5. Since all eigenvalues are distinct, corresponding eigenvectors X 1, X 2, X 3 are always linearly indepen-dent.

6. Dimensions of eigen spaces are, dim(Eλ1

)= 1, dim(Eλ2

)= 1, dim(Eλ3

)= 1.

Illustration 6.3 Find the eigenvalues and eigenvectors of the matrix

1 2 20 2 1

−1 2 2

. [Summer-2017]

Solution: Ifλ be the eigenvalue of A corresponding to eigenvector X ∈R3, then the characteristic equationis,

det(A−λI ) = 0 ⇒ λ3 −S1λ2 +S2λ−|A| = 0 (6.8)

where S1 =trace(A) = 1+2+2 = 5,

S2 =M11 +M22 +M33 =∣∣∣∣ 2 1

2 2

∣∣∣∣+ ∣∣∣∣ 1 2−1 2

∣∣∣∣+ ∣∣∣∣ 1 20 2

∣∣∣∣= 2+4+2 = 8,

|A| =∣∣∣∣∣∣

1 2 20 2 1−1 2 2

∣∣∣∣∣∣= 1(4−2)−2(0+1)+2(0+2) = 2−2+4 = 4.

Substituting in (6.8), we get

λ3 −5λ2 +8λ−4 = 0

∴ λ= 1, λ2 −4λ+4 = 0

∴ λ= 1, (λ−2)2 = 0

∴ λ= 1,2,2 =λ1,λ2,λ3(Say

)∴ Egien values of A are 1,2,2

(Repeated eigenvalues

)Now the eigenvector X 6= 0 ∈R3 is given by homogeneous system,

(A−λI ) X = 0 ⇒ 1−λ 2 2

0 2−λ 1−1 2 2−λ

xyz

=

0

0

0

(6.9)

λ===λ1 === 1 : From (6.9),

0 2 20 1 1−1 2 1

xyz

=

0

0

0

∴ From 2nd and 3rd equation,

y + z = 0, −x +2y + z = 0 ⇒ x = y =−z = t t ∈R





For t = 1, eigenvector corresponding to λ1 = 1 is X 1 = 1

1−1

.

λ===λ2 ===λ3 === 2 :(Note that for two equal eigenvalues, there may or may not exist two linearly independent eigenvectors.That is for two same eigenvalues, corresponding eigenvectors are may be two or one.)

From (6.9),

−1 2 20 0 1−1 2 0

xyz

= 0

00

∴ From 2nd and 3rd equation, z = 0, −x +2y = 0 ⇒ z = 0,

x

2= y = t , t ∈R

For t = 1, we get one eigenvector X 2 = 2

10

.

Observe that if we take any other value of t , we obtain an eigenvector which is constant multiple of X 2, thatis linearly dependent with X 2. Thus there exist only one linearly independent eigenvector correspondingtwo repeated eigenvalue λ2 =λ3 = 2. Hence,

λ2 =λ3 = 2 → X 2 = 2

10

Illustration 6.4 Find eigenvalues and eigenvectors of the matrix A = 5 0 1

1 1 0−7 1 0

.

Solution: eigenvalue of A is given by characteristic equation,

det(A−λI ) = 0

∴ λ3 −6λ2 +12λ−8 = 0 ⇒ (λ−2)(λ2 −4λ2 +4

)= 0

∴ (λ−2)3 = 0 ⇒ λ= 2,2,2(All eigenvalues are equal

)Corresponding to three equal eigenvalues, the linearly independent eigenvectors may be one or two orthree, given by

(A−λI ) X = 0,(

X 6= 0 ∈R3)

⇒ 5−λ 0 1

1 1−λ 0−7 1 0−λ

xyz

= 0

00

For λ= 2,

3 0 11 −1 0

−7 1 −2

xyz

= 0

00

From 1st and 2nd equation, 3x + z = 0, x − y = 0 ⇒ x = y =−z

3= t , t ∈R.

For t = 1, we have only one linearly independent eigenvector corresponding to three repeated eigenvalues

as X = 1

1−3


0 1 00 0 1

.

Solution: Given matrix is upper triangle matrix, so its are eigenvalues are its main diagonal elements. Thatis λ= 2,1,1 are eigenvalues of A. Now eigenvectors are given by

(A−λI ) X = 0,(

X 6= 0 ∈R3)

⇒ 2−λ 1 1

0 1−λ 00 0 1−λ

xyz

= 0

00





λ===λ1 === 2 :

0 1 10 −1 00 0 −1

xyz

= 0

00

.

From 2nd and 3rd equation, y = z = 0, x = t , t ∈R Hence, λ1 = 2 → X 1 = 1

00

λ===λ2 ===λ3 === 1 :

1 1 10 0 00 0 0

xyz

= 0

00

.

From 1st equation (2nd and 3rd columns are non pivot),

x + y + z = 0, y = t1, z = t2 ⇒ x =−t1,−t2, y = t1, z = t2, t1, t2 ∈RSince, for repeated eigenvalue (two same eigenvalues), we get two parametric solution. Hence we can findtwo linearly independent eigenvectors, by assuming the values of parameters t1, t2 as follow:

t1 = 1, t2 = 0 ⇒ X 2 = −1

10

, and t1 = 0, t2 = 1 ⇒ X 3 = −1

0−1

Thus two linearly independent eigenvectors corresponding to repeated eigenvalues like

λ2 =λ3 = 1 → X 2 = −1

10

, X 3 = −1

0−1

Illustration 6.6 Find the eigenvalue and eigenvector for the matrix A = 3 −1 0

−1 2 −10 −1 3

Solution: Here given matrix is symmetric. So it has always three linearly independent vectors irrespectiveof eigenvalues. The characteristic equation of A is

det(A−λI ) = 0

∴ λ3 −8λ2 +19λ−12 = 0 ⇒ λ= 1,3,4 =λ1,λ2,λ3

Now, eigenvectors are given by

(A−λI ) X = 0,(

X 6= 0 ∈R3)

⇒ 3−λ −1 0

−1 2−λ −10 −1 3−λ

xyz

= 0

00

λ===λ1 === 1 :

2 −1 0−1 1 −1

0 −1 2

xyz

= 0

00

From 1st and 3rd equation, 2x = y = 2z ∴ x = y

2= z = t , t ∈R. Hence, λ1 = 1 → X 1 =

121

λ===λ2 === 3 :

0 −1 0−1 −1 −1

0 −1 0

xyz

= 0

00

.

From 1st and 2nd equation, y = 0, x =−z = t , t ∈R. Hence, λ2 = 3 → X 1 = 1

0−1

λ===λ3 === 4 :

−1 −1 0−1 −2 −1

0 −1 −1

xyz

= 0

00

.





From 1st and 3rd equation, x =−y = z = t , t ∈R. Hence, λ3 = 4 → X 3 = 1

−1−1

Note: Since given matrix is symmetric its eigenvectors are always linearly independent and pair wise or-thogonal. That is X 1 ·X 2 = X 2 ·X 3X 3 ·X 1 = 0. (Verify !)


1 0 11 1 0

.

Solution: Given matrix is symmetric. Hence it has three linearly independent pair wise orthogonal eigen-vectors.Characteristic equation of A:

det(A−λI ) = 0 ⇒ λ3 −3λ−2 = 0 ⇒ λ= 2,−1,−1

eigenvectors are given by,

(A−λI ) X = 0,(

X 6= 0 ∈R3)

⇒ −λ 1 1

1 −λ 11 1 −λ

xyz

= 0

00

λ===λ1 === 2 :

−2 1 11 −2 11 1 −2

xyz

= 0

00

. From 1st and 2nd equation,

−2x + y + z = 0

x −2y + z = 0

⇒ x∣∣∣∣ 1 1

−2 1

∣∣∣∣=− y∣∣∣∣ −2 1

1 1

∣∣∣∣= z∣∣∣∣ −2 1

1 −2

∣∣∣∣[By Cramer’s rule

]

∴x

3= y

3= z

3⇒ x = y = z = t , t ∈R Hence, λ1 = 2 → X 1 =

111

λ===λ2 ===λ3 ===−−−1 :

1 1 11 1 11 1 1

xyz

= 0

00

.

∴ x + y + z = 0, y = t1, z = t2 ⇒ x =−t1 − t2, y = t1, z = t2, t1, t2 ∈RSince A is symmetric, it has thre linearly independent and pair wise orthogonal eigenvectors. Second vector

is given by taking t1 = 1 and t2 = 0 as X 2 = −1

10

.

â To find third vector X 3, we select t1 and t2 such that X 2 · X 3 = 0. This can be achieve by taking general

form of eigenvector X 3, that is X 3 = −t1 − t2

t1

t2

.

Now, X 2 ·X 3 = 0

∴ (−1,1,0) · (−t1 − t2, t1, t2) = 0

∴ (−1)(−t1 − t2)+ (1)(t1)+ (0)(t2) = 0

∴ 2t1 + t2 = 0 ⇒ t2 =−2t1

Thus for third vector if we take t1 = 1 then t2 =−2. Hence, X 3 = 1

1−2

.

∴ λ2 =λ3 =−1 → X 2 = −1

10

, X 3 = 1

1−2





6.5 Algebraic and Geometric Multiplicity of an eigenvalue

â The number of times an eigenvalue λ exist, is called an algebraic multiplicity of λ and is denoted bymulta(λ).

â The dimension of eigen space of λ is called geometric multiplicity of λ and is denoted by multg (λ).

Illustration 6.8 Determine the algebraic and geometric multiplicity of

0 1 11 0 11 1 0

. [Winter-2016]

Solution: To determine the algebraic and geometric multiplicity first find eigenvalues and eigenvectors.For given matrix we have obtained eigenvalues and eigenvectors in Illustration 6.7, and are

λ1 = 2 → X 1 = 1

11

λ2 =λ3 =−1 → X 2 = −1

10

, X 3 = 1

1−2

By definition,

1. eigenvalue λ= 2 exist one time and dimension of eigen space (number of corresponding eigenvector)is also one.

∴ Algebraic multiplicity of 2 = multa (2) = 1 and Geometric multiplicity of 2 = multg (2) = 1.

2. eigenvalue λ=−1 exist two time and dimension of eigen space (number of corresponding eigenvec-tor) is also two.

∴ Algebraic multiplicity of −1 = multa (−1) = 2 and Geometric multiplicity of −1 = multg (−1) = 2.

Note:

1. In Illustration 6.1, multa (4) = multa (9) = 1, multg (4) = multg (9) = 1.

2. In Illustration 6.2, multa (1) = multa (3) = multa (4) = 1, multg (1) = multg (3) = multg (4) = 1.

3. In Illustration 6.3, multa (1) = 1,multa (2) = 2, multg (1) = multg (2) = 1.

4. In Illustration 6.4, multa (2) = 3, multg (2) = 1.

5. In Illustration 6.5, multa (2) = 1, multg (1) = 2.

6. In Illustration 6.6, multa (1) = multa (3) = multa (4) = 1, multg (1) = multg (3) = multg (4) = 1.

Exercise 6.1

1. Find the eigenvalues, eigenvectors and hence the basis for the eigen space for the following matrices:

a.[

0 34 0

]b.

[1 00 1

]c.

[3 08 −1

]2. Non-symmetric matrix and non repeated eigenvalues:

a.

1 0 −11 2 12 2 3

[Winter-2015] b.

4 6 61 3 2−1 −4 −3

3. Non-symmetric matrix and repeated eigenvalues:

a.

1 0 02 0 13 1 0

b.

2 1 00 2 10 0 2

c.

4 6 61 3 2−1 −5 −2





4. Symmetric matrix and non repeated eigenvalues:

a.

−2 5 45 7 54 5 −2

b.

5 0 10 −2 01 0 5

5. Symmetric matrix and repeated eigenvalues:

a.

1 2 32 4 63 6 9

b.

3 1 11 3 −11 −1 3

6. Find the eigenvalues of A = −5 4 34

0 0 40 0 4

. Is A invertible ? [Summer-2016]

7. Determine the algebraic and geometric multiplicity of A = 0 1 0

0 0 11 −3 3

. [Winter-2015]

8. If λ is an eigenvalues of an orthogonal matrix A, prove that1

λis also an eigenvalue of A.

[Hint: For orthogonal matrix A−1 = AT ]

9. Let A be a 6×6 matrix with the characteristic equation λ2 (λ−1)(λ−2)3 = 0. What are the possibledimensions for eigen spaces for A ?

Answers

1. a.2p3

,− 2p3→

[ p3/21

],

[ −p3/21

]b. 1,1 →

[10

],

[01

]c. −1,3 →

[30

],

[04

]

2. a. 1,2,3 → −1

10

,

−212

,

−112

b. −1,1,4 → 6

2−7

,

0−11

,

31−1

3. a. −1 → 0

−11

, 1,1 → 0

11

b. 2,2,2 → 1

00

c. 1 → 4

1−3

, 2,2 → 3

1−2

4. a. −6,−3,12 → −1

01

,

1−1

1

,

121

b. 6,−2,4 → −1

01

,

010

,

101

5. a. 0,0,14 → −3

01

,

−15−3

,

123

b. 1,4,4 → −1

11

,

110

,

1−12

6. λ=−5,0,4, not invertible as λ= 0 is eigenvalue. 7. 1,1,1 → 1

11

, multa (1) = 3,multg (1) = 1

9. λ= 0 → dimEλ = 1 or 2, λ= 1 → dimEλ = 1, λ= 2 → dimEλ = 1 or 2 or 3.

E E E





6.6 Cayley-Hamilton Theorem 1

Every square matrix satisfies its own characteristic equation.

â For (2×2) matrix A, λ2 −S1λ+|A| = 0 ⇒ A2 −S1 A+|A| I2 = 0.

â For (3×3) matrix A, λ3 −S1λ2 +S2λ−|A| = 0 ⇒ A3 −S1 A2 +S A−|A| I3 = 0.

where S1 = trace(A) , S2 = M11 +M22 +M33, |A| = det(A) .

Illustration 6.9 Verify Cayley-Hamilton theorem for A =[

1 42 3

]. Hence find A3 and A−1.

Solution: The characteristic equation of given matrix is λ2 −5λ−2 = 0.∴ By Cayley-Hemilton theorem we have

A2 −5A−2I2 = 0 (6.10)

Verification:

A =[

1 32 4

]⇒ A2 = A A =

[1 32 4

][1 32 4

]=

[7 15

10 22

]∴ A2 −5A−2I2 =

[7 15

10 22

]−5

[1 32 4

]−2

[1 00 1

]=

[7 15

10 22

]−

[5 15

10 20

]−

[2 00 2

]=

[7−5−2 15−15−0

10−10−0 22−20−2

]=

[0 00 0

]= 0

Hence, A2 −5A−2I2 = 0 ∴ Cayley - Hemilton theorem is verified.

To find A3: Multiplying equation (6.10) by A, we get

A3 −5A2 −2A = 0 ⇒ A3 = 5A2 +2A

∴ A3 = 5

[7 15

10 22

]+2

[1 32 4

]=

[37 8154 118

]To find A−1: Multiplying equation (6.10) by A, we get

A−5I2 −2A−1 = 0 ⇒ A−1 = 1

2(A−5I2)

∴ A−1 = 1

2

([1 32 4

]−5

[1 00 1

])= 1

2

[ −4 32 −1

]∴ A−1 =

[ −2 3/21 −1/2

]

Illustration 6.10 Using Cayley-Hemilton theorem find A−1 if A = 2 1 1

0 1 01 1 2

. Hence find the matrix

represented by A8 −5A7 +7A6 −3A5 + A4 −5A3 +8A2 −2A+ I .

1Arthur Cayley; British, 1821-1895 and William Rowan Hamilton; Irish, 1805–1865.





Solution: Characteristic equation of given matrix is λ3 −5λ2 +7λ−3 = 0. Hence by Cayley-Hemilton the-orem

A3 −5A2 +7A−3I3 = 0 (6.11)

Multiplying both sides by A−1, we get

A2 −5A+7I3 −3A−1 = 0 ⇒ A−1 = 1

3

(A2 −5A+7I3

)∴ A−1 =1

3

2 1 10 1 01 1 2

2 1 10 1 01 1 2

−5

2 1 10 1 01 1 2

+5

1 0 00 1 00 0 1

=1

3

5 4 40 1 04 4 5

− 10 5 5

0 5 05 5 10

+ 5 0 0

0 5 00 0 5

∴ A−1 =1

3

2 −1 −10 3 0

−1 −1 2

= 2/3 −1/3 −1/3

0 1 0−1/3 −1/3 2/3

Now, to find matrix repented by A8 −5A7 +7A6 −3A5 + A4 −5A3 +8A2 −2A+ I , first we split this expressionwith one factor as LHS of equation (6.11), that is A3 −5A2 +5A− I3. This can be done using Long DivisionMethod for Polynomials (click here) as follow:

A8 −5A7 +7A6 −3A5 + A4 −5A3 +8A2 −2A+ I

A3 −5A2 +7A−3I= (

A5 + A)+ A2 + A+ I

A3 −5A2 +7A−3I, where I = I3

Multiply both the sides by(

A3 −5A2 +7A−3I), we get

A8 −5A7 +7A6 −3A5 + A4 −5A3 +8A2 −2A+ I =(A5 + A

)(A3 −5A2 +7A−3I

)+ (A2 + A+ I

)= (

A5 + A)

(0)+ (A2 + A+ I

)[∵ (6.11)]

= A2 + A+ I

= 5 4 4

0 1 04 4 5

+ 2 1 1

0 1 01 1 2

+ 1 0 0

0 1 00 0 1

∴ A8 −5A7 +7A6 −3A5 + A4 −5A3 +8A2 −2A+ I = 8 5 5

0 3 05 5 8

Exercise 6.2

1. Verify Cayley-Hamilton theorem for the following matrix A and hence find A3 and A−1:

a.[

1 23 4

][Winter-2015] b.

[ −1 13 0

]2. Verify Cayley-Hamilton theorem for the following matrix A and hence find A4 and A−1:

a.

2 −1 1−1 2 −1

1 −1 2

[Winter-2014, 2015] b.

6 −1 1−2 5 −1

2 1 7

[Summer-2017]

3. If A =[

1 42 3

]then simplify A5 −4A4 −7A3 +11A2 − A−10I .

4. Determine A−1 by using Cayley-Hamilton theorem for the matrix A = 1 3 2

0 −1 4−2 1 5

. Hence find

the matrix represented by A8 −5A7 − A6 +37A5 + A4 −5A3 −3A2 +41A+3I . [Winter-2016]


https://www.youtube.com/watch?v=jy4j-03j7pk&feature=youtu.be




Answers

1. a.[

37 5481 118

],

[ −2 13/2 −1/2

]b.

[ −7 412 −3

],

[0 1/31 1/3

]

2. a.

86 −85 85−85 86 −85

85 −85 86

,

3/4 1/4 −1/41/4 3/4 1/4

−1/4 1/4 3/4

b.

2176 −520 1400−1920 776 −1400

1920 520 2696

,

3/16 1/24 −1/481/16 5/24 1/48

−1/16 −1/24 7/48

3. A+5I =[

6 42 8

]

4.1

37

9 13 −148 −9 42 7 1

, −2A2 +4A+3I = 13 8 −40

16 −11 −1616 8 −27

E E E

6.7 Similar Matrices

Two matrices A and B are said to be similar if there exist a non-singular matrix P such that B = P−1 AP . Alsosimilar matrices have same eigenvalues.

6.8 Diagonalization

A matrix A is said to be diagonalizable (or can be diagonalizable) if it is similar to some diagonal matrix.That is there exist a non-singular matrix P such that P−1 AP = D , where D is a diagonal matrix.

Theorem 6.1 An n ×n matrix A is diagonalizable if and only if it has n linearly independent eigenvectors.In this case,

â A is similar to the diagonal matrix D = P−1 AP, where

â P is the matrix whose columns are the linearly independent eigenvectors and is known as ModalMatrix.

â D is the diagonal matrix whose diagonal elements are the eigenvalues of A and is known as a SpectralMatrix.

6.9 Orthogonally Diagonalization

â A matrix A is said to be orthogonally diagonalizable (or can be diagonalizable orthogonally) if thereexist an orthogonal matrix M such that M T AM = D , where D is a diagonal matrix.

â A matrix A is orthogonally diagonalizable if and only if it is symmetric.

â We know that an n ×n symmetric real matrix always has n linearly independent eigenvectors, eventhrough the eigenvalues are repeated.

â On normalizing each eigenvector we obtain the modal matrix M which is always orthogonal.

Illustration 6.11 Determine whether the following matrices are diagonalizable or not ? If so, diagonalizethem.

a. A = 2 0 −2

0 3 00 0 3

[Winter-2015] b. A = 1 2 1

2 0 −2−1 2 3





Solution: In order to digonalize, first of all we obtain eigenvalues and eigenvectors of given matrix.

a. Since given matrix is an upper triangle matrix, its eigenvalues are main diagonal elements. That isλ= 2,3,3 are eigenvalues of A.Now eigenvector is given by homogeneous system

det(A−λI ) X = 0,(

X 6= 0 ∈R3)

⇒ 2−λ 0 −2

0 3−λ 00 0 3−λ

xyz

= 0

00

λ===λ1 === 2 :

0 0 −20 1 00 0 1

xyz

= 0

00

From 1st and 2nd equation, y = z = 0, x = t , t ∈R. Hence, λ1 = 2 → X 1 =

100

.

λ===λ2 ===λ3 === 3 :

−1 0 −20 0 00 0 0

xyz

= 0

00

From 1st equation (2nd and 3rd columns are not pivot), x =−2z, y = t1, z = t2, t1, t2 ∈R.

Hence, λ2 =λ3 = 3 → X 2 = 0

10

, X 3 = −2

01

.

Since A has three linearly independent eigenvectors, it is diagonalizable. The Modal Matrix P which diago-nalize A is given by taking eigenvectors in columns, that is

P = [X 1 X 2 X 3

]= 1 0 −2

0 1 10 0 0

Also diagonalization of A is P−1 AP = D, where D = 2 0 0

0 3 00 0 3

= Spectral Matrix. [See Theorem 6.1]

b. Characteristic equation of A is,

λ3 −4λ2 +4λ= 0 ⇒ λ= 0,2,2

Eigen vectors are given by,

(A−λI ) X = 0,(

X 6= 0 ∈R3)

⇒ 1−λ 2 1

2 −λ −2−1 2 3−λ

xyz

= 0

00

λ===λ1 === 0 :

1 2 12 0 −2

−1 2 3

xyz

= 0

00

From 1st and 2nd equation, x +2y + z = 0, 2x −2z = 0 ⇒ x =−y = z = t , t ∈R.

Hence, λ1 = 0 → X 1 = 1

−11

.

λ===λ2 ===λ3 === 2 :

−1 2 12 −2 −2

−1 2 1

xyz

= 0

00

From 1st and 2nd equation, −x +2y + z = 0, 2x −2y −2z = 0 ⇒ y = 0, x = z = t , t ∈R.

Hence, λ2 =λ3 = 2 → X 2 = 1

01

.





Since given matrix A has only two linearly independent eigen vectors X 1, X 2 corresponding to three eigen-values λ= 0,2,2. Hence, it is not diagolalizable. [See Theorem 6.1]

Illustration 6.12 Find the normalized modal matrix M for the matrix A = 2 0 1

0 3 01 0 2

and diagonalize

orthogonally.

Solution: Observe that given matrix is symmetric because A = AT . Hence it is always orthogonally diago-nalizable. [See section 6.9]Characteristic equation of A : λ3 −7λ2 +15λ−9 = 0 ⇒ λ= 1,3,3.Eigen vectors are given by,

(A−λI ) X = 0,(

X 6= 0 ∈R3)

⇒ 2−λ 0 1

0 3−λ 01 0 2−λ

xyz

= 0

00

λ===λ1 === 1 :

1 0 10 2 01 0 1

xyz

= 0

00

From 1st and 2nd equation, y = 0, x =−z = t , t ∈R. ∴ λ1 = 1 → X 1 =

10−1

.

∴ Normalized eigenvector is, X1 = X 1∥∥∥X 1

∥∥∥ = 1p2

10−1

=

1/p

20

−1/p

2

λ===λ2 ===λ3 === 3 :

−1 0 10 0 01 0 −1

xyz

= 0

00

From 1st or 3rd equation, x = z = t1, y = t2, t1, t2 ∈ R. The first linearly independent vector is given bytaking t1 = 1, t2 = 0, that is

X 2 = 1

01

⇒ X2 = X 2∥∥∥X 2

∥∥∥ = 1p2

101

=

1/p

20

1/p

2

â Second linearly independent vector X 3 = t1

t2

t1

is such that [See Illustration 6.7]

X 2 ·X 3 = 0 ⇒ 2t1 = 0 ∴ t1 = 0.

So we can take any value of t2, let t2 = 1. X 3 = 0

10

⇒ X3 = X 3∥∥∥X 3

∥∥∥ = 0

10

. Thus, required nor-

malized modal matrix is

M = [X1 X2 X3

]= 1/

p2 1/

p2 0

0 0 1−1/

p2 1/

p2 0

Also diagnonalization of A is defined as,

M T AM = D = 1 0 0

0 3 00 0 3

= Spectral matrix.

Note:





1. Here the normalized modal matrix is an orthoganal matrix, that is M T = M−1.

2. Symmetric matrix is always orthoganally diagonalizable and hence diagonalizable.

3. For simply diadonalization, do not normalize the eigen vectors. In this case the modal matrix is

P = [X 1 X 2 X 3

]= 1 1 0

0 0 1−1 1 0

⇒ P−1 AP = D = 1 0 0

0 3 00 0 3

Exercise 6.3

1. For the following matrix, find the non singular matrix P and the diagonal matrix D such that D =P−1 AP.

a.[ −4 −6

3 5

][Winter-2016] b.

[5 33 5

]

c.

1 1 −2−1 2 10 1 −1

d.

1 1 31 5 13 1 1

2. Show that the matrix

1 1 10 1 10 0 1

can not be diagonalizable.

3. Find the normalized modal matrix M and diagonalize orthogonally the following matrices:

a.[

2 11 2

]b.

2 2 02 5 00 0 3

4. Prove that if b 6= 0, then

[a b0 a

]is not diagonalizable.

Answers

1. a. P =[ −2 −1

1 1

],D =

[ −1 00 2

]b. P =

[1 −11 1

],D =

[8 00 2

]

c. P = 1 1 3

0 3 21 1 1

,D = −1 0 0

0 2 00 0 1

d. P = 1 1 −1

−1 2 01 1 1

,D = 3 0 0

0 6 00 0 −2

3. a. P =[

−1/p

2 1/p

21/p

2 1/p

2

],D =

[1 00 3

]b. P =

0 −2/p

5 1/p

50 1/

p5 2/

p5

1 0 0

,D = 3 0 0

0 1 00 0 6

E E E

Powered by





Contact: (+91) 9427 80 9779





Chapter 7Quadratic Forms and Complex Matrices

7.1 Quadratic Form (QF)

â A homogeneous polynomial of degree two in n variables is called the quadratic form (QF) in n vari-ables.

â General QF in two variable:Q (x1, x2) = a11x2

1 +2a12x1x2 +a22x22

For example, 5x21 −2x2

2 +4x1x2, x2 −2x y are QF in two variables. But x2 −6y2 + x −5y is not a QFbecause all terms are not of degree two.

â General QF in three variable:

Q (x1, x2, x) = a11x21 +a22x2

2 +a33x23 +2a12x1x2 +2a23x2x3 +2a31x3x1

For example, 9x21−x2

2+4x23+6x1x2−8x1x3+x2x3, x1x2+x2x3+x3x1 are QF but x2

1−7x22+x2

3+4x1x2x3

is not QF as last term is of degree 3.

7.2 Matrix of Quadratic Form

Quadratic form in two and three variables can be expressed in matrix form as follow:

1. Q (x1, x2) = a11x21 +2a12x1x2 +a22x2

2 = X T AX ,

where X =[

x1

x2

]and A =

[a11 a12

a12 a22

].

2. Q (x1, x2, x3) = a11x21 +a22x2

2 +a33x23 +2a12x1x2 +2a23x2x3 +2a13x1x3 = X T AX ,

where X =

x1

x2

x3

and A = a11 a12 a13

a12 a22 a23

a13 a23 a33

.

* Important:

Observe that matrix form of a quadratic form is X T AX , where

â X is a column matrix of variable and

â A is a symmetric matrix in which diagonal entries are the coefficients of variables having square andother entries are half of the coefficients of cross multiplied variables, filled by symmetry in appropriatecolumns.

â In both representation A is a symmetric matrix. See the following illustration.

Illustration 7.1

78




a. 4x21 −9x2

2 −6x1x2 = X T AX , where X =[

x1

x2

], A =

[4 −3−3 −9

].

b.(x − y

)2 = x2 −2x y + y2 = X T AX , where X =[

xy

], A =

[1 −1−1 1

].

c. 3x21 +2x2

2 +3x23 −2x1x2 −2x3x2 = X T AX , where X =

x1

x2

x3

, A = 3 −1 0

−1 2 −10 −1 3

.

d. 2x2 +5y2 −6z2 −2x y − y z +8xz = X T AX , where X = x

yz

, A = 2 −1 −1/2

−1 5 4−1/2 4 −6

.

[summer-2015]

7.3 Index, Signature and Rank of Quadratic Form

For quadratic form X T AX ,

â Number of positive eigenvalues of A is called index of QF.

â The difference between number of positive and negative eigenvalues of A is called signature of QF.

â Number of non zero eigen value is called rank of QF.

7.4 Definiteness of Quadratic Form

A quadratic form Q (x1, x2, x3.....xn) = X T AX is said to be

1. Positive definite if all eigenvalues of A are positive.

2. Negative definite if all eigenvalues of A are negative.

3. Semi-positive definite if all eigenvalues are positive and atleast one eigenvalue of A is zero.

4. Semi-negative definite if all eigenvalues are negative and atleast one eigenvalue is zero.

5. Infinite if some eigenvalues of A are positive and some eigenvalues of A are negative.

Illustration 7.2 Determine the index, signature, rank and definiteness of the quadratic form −3x2−5y2−3z2 +2x y +2y z −2xz.

Solution: Matrix form of quadratic form is

−3x2 −5y2 −3z2 +2x y +2y z −2xz = X T AX , where X = x

yz

, A = −3 1 1

1 −5 −11 −1 −3

The characteristic equation of A is

λ3 +11λ2 +36λ+36 = 0 ⇒ λ=−2,−3,−6

1. Index = Number of positive eigenvalues = 0.

2. Signature = Difference of number of positive and number of negative eigen values.

Here A has no positive eigenvalue, so number of positive eigen value is 0. Also A has all three negativeeigenvalues, so number of negative eigenvalues is 3.

Hence difference between +ve and −ve eigenvalues is 3. (always consider difference in modulus)

∴ Signature = 3.





3. Rank = Number of non zero eigen value = 3.

4. Since all eigen values of A are non zero and negative.

Hence given quadratic form is of negative definite.

Theorem 7.1 (Principal Axes Theorem)Let Q (x1, x2, x3.....xn) = X T AX be a quadratic form where A is an n×n symmetric matrix. Then Q (x1, x2, x3.....xn)can be transformed into λ1 y2

1 +λ2 y22 +λ3 y2

3 + .....+λn y2n by the orthogonal linear transformation X = PY ,

where P is an orthogonal modal matrix of A and λ1,λ2,λ3, .....λn are the eigenvalues of A.â The reduced form of QF is known as canonical form (or sum of squares) of Q (x1, x2, x3.....xn).

Illustration 7.3 Find the canonical form of the quadratic form 2x21 +3x2

2 +2x23 +2x1x3, using orthogonal

transformation. Also find index, rank and signature of the quadratic form. [Summer-2014]

Solution: Matrix form of given quadratic form is,

2x21 +3x2

2 +2x23 +2x1x3 = X T AX , where X =

x1

x2

x3

, A = 3 0 1

0 3 01 0 2

Characteristic equation of A :

λ3 −8λ2 +20λ−15 = 0 ⇒ (λ−3)(λ2 −5λ+5

)= 0

∴ λ= 3, λ2 −5λ+5 = 0

∴ λ= 3, λ= 5±p25−20

2

[∵ ax2 +bx + c = 0 ⇒ x = −b ±

pb2 −4ac

2a

]

∴ λ= 3,5+p

5

2,

5−p5

2=λ1,λ2,λ3

Thus by Principal Axis Theorem [Theorem 7.1], given quadratic form can be reduce to canonical form, underthe orthog0nal transformation X = PY as

X T AX =λ1 y21 +λ2 y2

2 +λ3 y23 = 3y2

1 +(

5+p5

2

)y2

2 +(

5−p5

2

)y2

3

where P is normalized modal matrix of A, X = x1

x2

x3

and Y = y1

y2

y3

. Also,

1. Index = Number of +ve eigenvalue = 3.

2. Rank = Number of non zero eigenvalue = 3.

3. Signature = Difference of +ve and −ve eigenvalue = 3.

Note: In order to find orthogonal transformation X = PY that reduce given quadratic form to canonicalform (to verify principal axis theorem) it is essential to find normalized modal matrix P of symmetric matrixA. See below illustration.

Illustration 7.4 Determine the orthogonal transformation which transform the quadratic form 5x2+5y2+5z2 +4x y +4y z +4zx into canonical form.

Solution: Given quadratic form

5x21 +5x2

2 +5x23 +4x1x2 +4x2x3 +4x3x1 = X T AX , X =

x1

x2

x3

, A = 5 2 2

2 5 22 2 5





∴ Characteristic equation of A : λ3 −15λ2 +63λ−81 = 0 ⇒ λ= 3,3,9. Now eigenvectors are given by

(A−λI ) X = 0,(

X 6= 0 ∈R3)

⇒ 5−λ 2 2

2 5−λ 22 2 5−λ

xyz

= 0

00

λ===λ1 ===λ2 === 3 :

2 2 22 2 22 2 2

xyz

= 0

00

This yields, x + y + z = 0 ⇒ x =−t1 − t2, y = t1, z = t2, t1, t2 ∈R.∴ Two linearly independent vectors are

X 1 = −1

10

⇒ X1 = X 1∥∥∥X 1

∥∥∥ = 1p2

−110

=

−1/p

21/p

20

and

X 2 = 1

1−2

⇒ X2 = X 2∥∥∥X 2

∥∥∥ = 1p6

11−2

=

1/p

61/p

6−2/

p6

λ===λ1 ===λ2 === 3 :

−4 2 22 −4 22 2 −4

xyz

= 0

00

∴ From 1st and 2nd equation, −2x + y + z = 0, x −2y + z = 0 ⇒ x = y = z = t , t ∈R.

∴ X 3 = 1

11

⇒ X3 = X 3∥∥∥X 3

∥∥∥ = 1p3

111

=

1/p

31/p

31/p

3

The normalized modal matrix of A is P = [X 1 X 2 X 3

]= −1/

p2 1/

p6 1/

p3

1/p

2 1/p

6 1/p

30 −2/

p6 1/

p3

.

Hence required orthogonal transformation is,

X = PY ⇒ x1

x2

x3

=

−1/p

2 1/p

6 1/p

31/p

2 1/p

6 1/p

30 −2/

p6 1/

p3

y1

y2

y3

∴ x1 =− y1p

2+ y2p

6+ y3p

3, x2 = y1p

2+ y2p

6+ y3p

3, x3 =−2

y2p6+ y3p

3

â Note that, if we substitute these values of x1, x2, x3 in given quadratic form and simplify, we get thecanonical form of quadratic form as X T AX = 3y2

1 +3y22 +9y2

3 . This is the statement principal axis theorem.(Verify !)Note: Recall from following table, some quadratic equations and its geometrical nature/name:

Equation Nature/Name

1. x2 + y2 = a2 Circle

2.x2

a2 + y2

b2 = 1 Ellipse

3.x2

a2 − y2

b2 = 1 Hyperbola





4. x2 + y2 + z2 = a2 Sphere

5.x2

a2 + y2

b2 + z2

c2 = 1 Ellipsoid

6.x2

a2 ± y2

b2 ∓ z2

c2 = 1 orx2

a2 − y2

b2 − z2

c2 = 1 Hyperboloid

Illustration 7.5 Find the nature of the graph represented by the following equations:

a. x2 +4x y + y2 = 16 b. 5x2 +5y2 +5z2 +4x y +4xz +4y z = 9

Solution:

a. The quadratic form corresponding to given eqution is

x2 +4x y + y2 = X T AX , where X =[

xy

], A =

[1 22 1

]Now characteristic equation of A : λ2 −2λ−3 = 0 ⇒ λ= 3,−1 =λ1,λ2

Hence by principal axis theorem, using orthogonal transformation given quadratic form will be trans-formed in to canonical form as

X T AX =λ1 y21 +λ2 y2

2 ⇒ x2 +4x y + y2 = 3y21 − y2

2 = 16 [∵ Given]

∴ 3y21 − y2

2 = 16 ⇒ y21

16/3− y2

2

16= 1 → Hyperbola [See 3rd equation in above table]

∴ Given quadratic equation represent the curve hyperbola.

b. The eigenvalues for given quadratic form are λ= 3,3,9. [See Illustration 7.4]∴ By principal axis theorem, we have

5x2 +5y2 +5z2 +4x y +4xz +4y z = 3y21 +3y2

2 +9y23 = 9

∴ 3y21 +3y2

2 +9y23 = 9

⇒ y21

3+ y2

2

3+ y2

3

1= 1 → Ellipsoid [See 5th equation in above table]

∴ Given quadratic equation represent the surface of ellipsoid.

Exercise 7.1

1. Which of the following forms are the quadratic form ? If so, express it as matrix form X T AX anddetermine the its index, signature, rank and definiteness.

a. x2 −2x y b. 3x21 +7x2

2

c. x y + y z + zx d. 4x21 +x2

2 +15x23 −4x1x2

2

2. Reduce the following quadratic form to the canonical form (sum of square) by using orthogonal lineartransformation and write the rank, index and signature:

a. 2(x2

1 +x1x2 +x22

)b. 2x2

1 +5x22 +3x2

3 +4x1x2

c. 2x21 +x2

2 −3x23 d. 3x2 +3z2 +8x y +8xz +8y z [Winter-2014]

3. Find the nature of the graph represented by the following equations: (Name the quadratic)

a. x2 +4x y +3y2 = 4 b. 2x2 −4x y +2y2 = 1

c. 5x2 −4x y +8y2 −36 = 0 d. 5x2 −2y2 +5z2 +2xz = 1





Answers

1. a. Index 1, Signature 0, Rank 2, Infinite form. b. Index 2, Signature 2, Rank 2, Positive definiteform. c. Index 1, Signature 1, Rank 3, Infinite form. d. not quadratic form

2. a. 3y21 + y2

2 , Rank 2, Index 2, Signature 2 b. y21 +3y2

2 +6y23 , Rank 3, index 3, Signature 3

c. 2y21 + y2

2 −3y23 , Rank 2, Index 2, Signature 1

d. −y21 +

1

2

(7−p

177)

y22 +

1

2

(7+p

177)

y23 , Rank 3, Index 1, Signature 1

3. a. Hyperbola b. Ellipse c. Ellipse d. Hyperboloid

E E E

7.5 Complex Matrix

A matrix is said to be complex matrix if it has at least one complex entry otherwise it is known as real matrix.

e. g.

A =[

1 2+ i−i 5

], A =

1 2 −10 i 33 7 4

, where i 2 =−1

7.6 Conjugate Matrix

Matrix obtained by replacing the elements of a complex matrix A by its complex conjugate numbers is saidto be conjugate matrix of A and is denoted by A.e. g.

A =[

1 2+ i−i 5

]⇒ A =

[1 2− ii 5

]

7.7 Conjugate Transpose

The conjugate transpose of a complex matrix A is denoted A∗ and is define as conjugate of transpose (or

transpose of conjugate) of A. That is (A′) =(

A)′ = A∗.

e. g.

A = 1 i 3+2i

0 2 −4i−3 1− i 5

⇒ A∗ =(

A)T =

1 −i 3−2i0 2 4i−3 1+ i 5

T

= 1 0 −3

−i 2 1+ i3−2i 4i 5

7.8 Hermitian, Skew-Hermitian, Unitary and Normal Matrices

A square matrix A = [ai j

]of size n ×n is said to be

1. Hermitian if A∗ = A, that is ai j = a j i . (Elements of main diagonal are purely real)

2. Hermitian if A∗ =−A, that is ai j =−a j i . (Elements of main diagonal are purely imaginary)

3. Unitary if A∗ = A−1 that is A∗A = A A∗ = In .

4. Normal if A∗A = A A∗





* Important:

â Observe that the real symmetric and skew-symmetric matrices are complex analogous of Hermitionand Skew-Hermition matrices.

â Every Hermitian matrix is normal since A∗A = A A = A A∗ and every unitary matrix is normal matrixsince A∗A = I = A A∗

â Eigenvalues of Hermitian matrix are real.

â Eigenvectors of normal matrix A corresponding to different eigen spaces are orthogonal.

7.9 Properties

1.(

A∗)∗ = A 2. (A±B)∗ = A∗±B∗

3. (k A)∗ = k A∗ 4. (AB)∗ = B∗A∗

5.(

A)= A 6. AB = A B

7. (k A) = k A 8. (A±B) = A±B

9. det(

A)= det(A)

Illustration 7.6 Find k, l and m to make A a Hermitian matrix; where A = −1 k −i

3−5i 0 ml 2+4i 2

.

Solution: We know that A is Hermitian if A∗ = A, that is ai j = a j i . Thus for given matrix,

k = a12 = a21 = (3−5i ) = 3+5i ∴ k = 3+5i

l = a31 = a13 = (−i ) = i ∴ l = i

m = a23 = a32 = (2+4i ) = 2− i 4 ∴ m = 2− i 4

Illustration 7.7 Prove that det(

A∗)= det(A).

Solution: We know that, A∗ =(

A)T

⇒ det(

A∗)= det

[(A

)T]

= det(

A) [

∵ det A = det AT ]= det A

[∵ Property 9

]∴ det

(A∗)= det A Proved.

Exercise 7.2

1. In each part find A∗:

a. A = 2i 1− i

4 3+ i5+ i 0

b. A = 2i 1− i −1+ i

4 5−7i −ii 3 1

2. Which of the following are Hermitian matrices ?

a. A =[

1 1+ i1− i −3

]b. A =

[i i−i i

]





c. A = −2 1− i −1+ i

1+ i 0 3−1− i 3 5

d. A = 1 0 0

0 1 00 0 1

[Hint: For Hermitian matrix A∗ = A.]

3. Prove that,

a. If A is Hermitian then det(A) is real.

b. If A is Unitary then |det(A) | = 1.

c. The entries of main diagonal of the Hermitian matrix are real numbers.

d. If A is Unitary then A∗is also unitary.

Answers

1. a.[ −2i 4 5− i

1+ i 3− i 0

]b.

−2i 4 −i1+ i 5+7i 3−1− i i 1

2. a, c, d yes, b no.

E E E

Powered by





Contact: (+91) 9427 80 9779





Chapter 8Inner Product Space and Orthogonal Basis

8.1 Inner Product Space

Let V be the real vector space then a mapping ⟨ ,⟩ : V ×V →R is said to be an inner product on V if it satisfiesthe following axioms:

∀ u, v , w ∈V and α ∈R1.

⟨u, v

⟩= ⟨v ,u

⟩[Symmetry]

2.⟨

u + v , w⟩= ⟨

u, w⟩+⟨

v , w⟩

[Additive]

3.⟨αu, v

⟩=α⟨u, v

⟩[Homogeneity]

4.⟨

u,u⟩Ê 0 [Positivity]

5.⟨

u,u⟩= 0 ⇔ u = 0

A vector space together with an inner product is called an inner product space.

8.2 Properties of Inner Product

Let V be the real inner product space. For u, v , w ∈V and α ∈R1.

⟨0,u

⟩= ⟨u,0

⟩= 0

2.⟨

u, v +w⟩= ⟨

u, v⟩+⟨

u, w⟩

3.⟨

u,αv⟩=α⟨

u, v⟩= 0

4.⟨

u − v ,+w⟩= ⟨

u, w⟩−⟨

v , w⟩

5.⟨

u, v −w⟩= ⟨

u, v⟩−⟨

u, w⟩

8.3 Some Standard Inner Product Spaces

1. Euclidean inner product space Rn :

Let u = (u1,u2,u3.....un) and v = (v1, v2, v3.....vn) are vectors of Rn then the formula⟨u, v

⟩= u · v = u1v1 +u2v2 +u3v3 + .....+un vn

defines inner product on Rn and hence Rn is an inner product space.

Here the inner product, define as above is known as standard inner product on Rn .e. g. Let u = (1,−1,2) , v = (2,1,3) ∈R3 ⇒ ⟨

u, v⟩= u · v = 2−1+6 = 7.

86




2. Weighted inner product space Rn : Let u = (u1,u2,u3.....un) and v = (v1, v2, v3.....vn) are vectors ofRn and w1, w2, w3.....wn are positive numbers which we shall call weight, then the weighted innerproduct is defined as ⟨

u, v⟩= w1u1v1 +w2u2v2 +w3u3v3 + .....+wnun vn

e. g. u = (1,2,−2) , v = (1,1,3) ∈R3 w1 = 2, w2 = 1

2, w3 = 1

⇒ ⟨u, v

⟩= 2(1)(1)+ 1

2(2)(1)+1(−2)(3) =−3.

3. An Inner product generated by matrix:

Let u, v ∈Rn and A be invertible n ×n then an inner product generated by matrix A is defined by,⟨u, v

⟩= Au · Av

e. g. Let u = (1,2) , v = (2,−3) ∈R2, A =[

1 −14 2

]⇒ Au =

[1 −14 2

][12

]=

[ −18

]= (−1,8) , Av =

[1 −14 2

][2−3

]=

[52

]= (5,2)

∴⟨

u, v⟩= Au · Av = (−1,8) · (5,2) =−5+16 = 11

∴⟨

u, v⟩= 11

4. Inner product on M 22:

Let A =[

a1 a2

a3 a4

],B =

[b1 b2

b3 b4

]∈M22, then the standard inner product on M22 is defined by

⟨A,B⟩ = trace(B T A

)a1b1 +a2b2 +a3b3 +a4b4

e. g. Let A =[

1 34 2

], B =

[ −2 30 5

]∈M22 ⇒ ⟨A,B⟩ =−2+9+0+10 = 17.

5. Inner product on P 2(x) : Let p = a0 + a1x + a2x2, q = b0 +b1x +b2x2 ∈ P2 (x). Then the standardinner product on P2(x) is defined by⟨

p, q⟩= a0b0 +a1b1 +a2b2

Similarly, we can extend the definition on Pn(x).e. g. Let p = 1+2x +x2, q = 2−4x +5x2 ∈P2 (x) ⇒ 2−8+5 =−1.

6. Inner product on C [a, b]:

Let f = f (x) , g = g (x) ∈C [a,b] (Set of all continuous functions defined on [a,b]). The standard innerproduct on C [a,b] is defined by ⟨

f , g⟩= ∫ b

af (x) g (x)d x

e. g. Let f (x) = x +x2, g (x) =−x ∈C [−1,1]

⇒ ⟨f , g

⟩= ∫ 1

−1f (x) g (x)d x =

∫ 1

−1

(−x2 −x3)d x

=−[

x3

3+ x4

4

]1

−1=−

[(1

3+ 1

4

)−

(−1

3+ 1

4

)]∴

⟨f , g

⟩=−2

3





Illustration 8.1 Let u = (u1,u2) and v = (v1, v2) be vectors in R2. Verify that the weighted Euclidean innerproduct

⟨u, v

⟩= 3u1v1 +2u2v2 satisfies the inner product axioms. [Winter-2017]

Solution: Let u = (u1,u2) , v = (v1, v2) , w = (w1, w2) ∈R2 and α ∈R.

1.⟨

u, v⟩= 3u1v1 +2u2v2

= 3v1u1 +2v2u2

∴⟨

u, v⟩= ⟨

v ,u⟩

2.⟨

u + v , w⟩= 3(u1 + v1) w1 +2(u2 + v2) w2

= 3(u1w1 + v1w1)+2(u2w2 + v2w2)

= (3u1w1 +2u2w2)+ (3v1w1 +2v2w2)

∴⟨

u + v , w⟩= ⟨

u, w⟩+⟨

v , w⟩

3.⟨αu, v

⟩= 3(αu1) v1 +2(αu2) v2

=α (3u1v1 +2u2v2)

∴⟨

u, v⟩=α⟨

u, v⟩

4.⟨

u,u⟩= 3u1u1 +2u2u2

= 3u21 +2u2

2 Ê 0

∴⟨

u, v⟩Ê 0

5.⟨

u,u⟩= 0 ⇔ 3u1u1 +2u2u2 = 0

⇔ 3u21 +2u2

2 = 0

⇔ u1 = 2u2 = 0

⇔ u = (u1,u2) = (0,0) = 0

∴⟨

u, v⟩⇔ u == 0

Hence, given product satisfies all the inner product axioms.

8.4 Norm, Distance and Angle

Let V be the real inner product space and u, v ∈V , then norm, distance and angle are defined as

1. Norm of vector:∥∥u

∥∥=√⟨

u,u⟩

2. Distance between two vectors : d(u, v

)= ∥∥u − v∥∥=

√⟨u − v ,u − v

⟩3. Angle θ between two vectors: cosθ =

⟨u, v

⟩∥∥u∥∥∥∥v

∥∥Also u and v are said to be orthogonal to each other if

⟨u, v

⟩= 0(i.e. θ = 90) and is denoted by u⊥v .

Illustration 8.2 Let R4 have the Euclidean inner product. Find the cosine of the angle θ and distancebetween the vectors u = (4,3,2,−1) and v = (−2,1,2,3) . [Winter-2017]

Solution: Given u = (4,3,2,−1) , v = (−2,1,2,3) . With respect to standard inner product in R4, we have⟨u, v

⟩= u · v =−8+3+4−3 =−4∥∥u∥∥=p

16+9+4+1 =p30,

∥∥v∥∥=p

4+1+4+9 =p18 = 3

p2

Angle the cosine of between two vectors is defined by

cosθ =⟨

u, v⟩∥∥u

∥∥∥∥v∥∥ = u · v∥∥u

∥∥∥∥v∥∥ = −4(p

30)(

3p

2) =− 4

6p

15∴ cosθ =− 2

3p

15





Also distance between two vectors is given by,

d(u, v

)= ∥∥u − v∥∥= ‖(4,3,2,−1)− (−2,1,2,3)‖ = ‖(6,2,0,−4)‖

∴ d(u, v

)=p36+4+0+16 =p

56 = 3p

6 ∴ d(u, v

)= 3p

6

Illustration 8.3 Find∥∥p

∥∥, d(p, q

)and cosθ using standard inner product on P2 where p = −3 − x +

x2, q = 2+x2.

Solution: Standard inner product in P2 is defined as⟨

p, q⟩= a0b0 +a1b1 +a2b2.

â Norm of p is∥∥p∥∥=

√⟨p, p

⟩=pa0a0 +a1a1 +a2a2 =

√a2

0 +a21 +a2

2

∴∥∥p

∥∥=√

(−3)2 + (−1)2 + (1)2 =p11

[∵ p =−3−x +x2]

â Distance between p and q is

d(p, q

)= ∥∥p −q∥∥= ∥∥(−3−x +x2)− (

2+x2)∥∥= ∥∥(−5−x2)∥∥∴ d

(p, q

)=√(−5)2 + (0)2 + (1)2 =p

26

â Cosine of angle θ between p and q is

cosθ =⟨

p, q⟩∥∥p

∥∥∥∥q∥∥ = (−3)(2)+ (−1)(0)+ (1)(1)(√

(−3)2 + (−1)2 + (1)2)(√

(2)2 + (0)2 + (1)2)

∴ cosθ = −6+0+1p11

p5

=− 5p55

8.5 Results

1.∥∥u

∥∥Ê 0 2.∥∥αu

∥∥= |α |∥∥u∥∥

3. Cauchy-Schwarz’s inequality: Let V be the real inner product space and u, v ∈V then∣∣⟨u, v⟩∣∣É ∥∥u

∥∥ ∥∥v∥∥

Proof: Angle between two vectors of an inner product space V is defined by,

cosθ =⟨

u, v⟩∥∥u

∥∥∥∥v∥∥

Since |cosθ| É 1 ⇒∣∣⟨u, v

⟩∣∣∥∥u∥∥∥∥v

∥∥ É 1 ∴∣∣⟨u, v

⟩∣∣É ∥∥u∥∥∥∥v

∥∥ Proved.

4. Triangle inequality: Let V be the real inner product space and u, v ∈V then∥∥u + v∥∥É ∥∥u

∥∥ + ∥∥v∥∥

Proof:∥∥u + v

∥∥2 = ⟨u + v ,u + v

⟩ [∵By definition of norm

]= ⟨

u,u + v⟩+⟨

v ,u + v⟩ [

∵By definition of inner product]

= ⟨u,u

⟩+⟨u, v

⟩+⟨v ,u

⟩+⟨v , v

⟩= ∥∥u

∥∥2 +2⟨

u, v⟩+∥∥v

∥∥2 [∵

⟨u, v

⟩= ⟨v ,u

⟩]É ∥∥u

∥∥2 +2∣∣⟨u, v

⟩∣∣+∥∥v∥∥2 [

∵⟨

u, v⟩É ∣∣⟨u, v

⟩∣∣]É ∥∥u

∥∥2 +2∥∥u

∥∥∥∥v∥∥+∥∥v

∥∥2 [∵Cauchy - Schwarz’s inequality

]∥∥u + v∥∥2 É (∥∥u

∥∥+∥∥v∥∥)2

∴∥∥u + v

∥∥É ∥∥u∥∥+∥∥v

∥∥ Proved.





5. Generalized Pythagoras Theorem: If u and v are orthogonal vectors of the real inner product spaceV then ∥∥u + v

∥∥2 = ∥∥u∥∥2 +∥∥v

∥∥2

Proof: Since u and v are orthogonal vectors, we have⟨

u, v⟩= 0.

Now∥∥u + v

∥∥2 = ⟨u + v ,u + v

⟩= ⟨

u,u + v⟩+⟨

v ,u + v⟩

= ⟨u,u

⟩+⟨u, v

⟩+⟨v ,u

⟩+⟨v , v

⟩= ∥∥u

∥∥2 +0+0+∥∥v∥∥2 [

∵⟨

u, v⟩= ⟨

v ,u⟩= 0

]∴

∥∥u + v∥∥2 É ∥∥u

∥∥2 +∥∥v∥∥2 Proved.

Illustration 8.4 Verify Cauchy-Schwarz’s inequality for u = (−2,1) and v = (1,0) , using the inner product⟨u, v

⟩= 4u1v1 +5u2v2.

Solution: For given weighted ineer product⟨

u, v⟩= 4u1v1 +5u2v2,∥∥u

∥∥=√⟨

u,u⟩=√

4u1u1 +5u2u2 =√

4u21 +5u2

2

⇒ ∥∥u∥∥=

√4(−2)2 +5(1)1 =p

21[∵ u = (−2,1)

]Similarly,

∥∥v∥∥=

√4(1)2 +5(0) = 2

[∵ v = (1,0)

]∴

∥∥u∥∥∥∥v

∥∥= 2p

21 (8.1)

Also,⟨

u, v⟩= 4u1v1 +5u2v2 = 4(−2)(1)+5(1)(0) =−8

∴∣∣⟨u, v

⟩∣∣= 8 (8.2)

∴ From (8.1) and (8.2), we have∣∣⟨u, v

⟩∣∣É ∥∥u∥∥∥∥v

∥∥Hence, Cauchy-Schwarz’s inequality is satisfied.

Illustration 8.5 If u and v are unit vectors such that⟨

u, v⟩=−1, evaluate

∥∥2u − v∥∥ .

Solution: Using definitions of inner product and norm,∥∥2u − v∥∥2 = ⟨

2u − v ,2u − v⟩

= ⟨2u − v ,2u

⟩+⟨2u − v ,−v

⟩= ⟨

2u ,2u⟩+⟨−v ,2u

⟩+⟨2u,−v

⟩+⟨−v ,−v⟩

= 4⟨

u ,u⟩−2

⟨v ,u

⟩−2⟨

u , v⟩+⟨

v , v⟩

[∵Axiom 3 of definition]

= 4⟨

u ,u⟩−4

⟨u , v

⟩+⟨v , v

⟩ [∵

⟨v ,u

⟩= ⟨u , v

⟩]= 4

∥∥u∥∥2 −4

⟨u , v

⟩+∥∥v∥∥2

[∵Definition of norm]

= 4(1)−4(−1)+ (1) [∵Given]∥∥2u − v∥∥2 = 9 ⇒ ∥∥2u − v

∥∥= 3

Illustration 8.6 Use Cauchy-Schwarz inequality to prove for all real values of a,b,θ,

(a cosθ+b sinθ)2 É a2 +b2

Solution: Let u = (a,b) , v = (cosθ, sinθ) ∈R2.∴ By cauchy-Schwarz inequality in R2, with standard inner product (dot product), we have∣∣⟨u, v

⟩∣∣É ∥∥u∥∥∥∥v

∥∥ ⇒ ∣∣u · v∣∣É ∥∥u

∥∥∥∥v∥∥

∴ |a cosθ+b sinθ| É√

a2 +b2√

cos2θ+ sin2θ =√

a2 +b2p

1

∴ |a cosθ+b sinθ| É√

a2 +b2

Hence, (a cosθ+b sinθ)2 É a2 +b2 [∵ Squaring both the sides

]





Exercise 8.1

1. Show that⟨

u, v⟩ = 9u1v1 + 4u2v2 is the inner product on R2 generated by the matrix A =

[3 00 2

].

Also find⟨

u, v⟩

for u = (−3,2) and v = (1,7) .

2. For u = (u1,u2,u3) , v = (v1, v2, v3) ∈R3, show that⟨

u, v⟩= u2

1v21 +u2

2v22 +u2

3v23 . is not an inner product.

3. Let p = p (x) , q = q (x) ∈ P2. Show that⟨

p, q⟩ = p (0) q (0)+p

(12

)q

(12

)+p (1) q (1) is an inner producton P2. Is this inner product on P3 ? Explain.

4. In each part use the given inner product on R2 to find∥∥w

∥∥ and d(u, v

), where w = (−1,3) ,u =

(−1,2) , v = (2,5).

a. The standard Euclidean inner product.

b. The weighted inner product⟨

u, v⟩= 3u1v1 +2u2v2.

c. The inner product generated by A =(

1 2−1 1

).

5. For M22 find ‖ A‖ and d (A,B) given that, A =[

2 69 4

],B =

[ −4 71 6

]6. In each part, verify the Cauchy-Schwarz inequality:

a. p =−1+2x +x2, q = 2x using standard inner product in P2.

b. U =( −1 2

6 1

), V =

(1 03 3

)using inner product ⟨U ,V ⟩ = trace

(VTU

).

7. Prove that∥∥u + v

∥∥2 +∥∥u − v∥∥2 = 2

∥∥u∥∥2 +2

∥∥v∥∥2.

8. If u and v are an (n ×1) matrices and A be an (n ×n) matrix then, prove that(vT AT Au

)2 É(uT AT Au

)(vT AT Av

)[Hint: Use Cauchy-Schwarz inequality for inner product

⟨u, v

⟩= Au · Av = vT AT Au]

9. Show that, equality holds in Cauchy-Schwarz inequality if and only if u and v are linearly dependent.

[Hint: u, v L.D. ⇔ u = kv ⇔ θ = 0 ⇔ cosθ = 1 ⇔∣∣⟨u, v

⟩∣∣∥∥u∥∥∥∥v

∥∥ = 1 ⇔ ∣∣⟨u, v⟩∣∣= ∥∥u

∥∥∥∥v∥∥]

10. If⟨

u, v⟩= 2,

⟨v , w

⟩=−3,⟨

u, w⟩= 5,

∥∥u∥∥= 1,

∥∥v∥∥= 2,

∥∥w∥∥= 7, evaluate the following expressions:

a.⟨

u − v −2w ,4u + v⟩

b.∥∥u −2v +4w

∥∥11. Prove that

⟨u, v

⟩= 1

4

∥∥u + v∥∥2 − 1

4

∥∥u − v∥∥2.

12. With respect to the Euclidean inner product, the vectors u =(1,p

3)

and v =(−1,

p3)

have norm 2,

and the angle between them is 60. Find the weighted Euclidean inner product with respect to whichu and v are orthogonal unit vectors.

Answers

8.6 Orthogonal Complement

â Let W be a subspace of an inner product space V . A vector u ∈V is said to be orthogonal to W if it isorthogonal to every vector of W .

â The set of all vectors of V that are orthogonal to W is called the orthogonal complement of W and isdenoted by W ⊥. That is,

W ⊥ = u ∈V :

⟨u, w

⟩= 0,∀w ∈W





8.7 Properties of W⊥

1. W ⊥ is a sunspace of the inner space V. 2. The only common vector in W and W ⊥ is 0.

3.(

W ⊥)⊥ =W =W ⊥⊥ 4. dimW +dimW ⊥ = dimV

5. Let W be a subspace of an inner product space V. Then u ∈ W ⊥ (that is u is orthogonal to W ) if andonly if u is orthogonal to every vectors of spanning set of W .

Thus, if W = span

w1, w2, w3, ...wn

then u ∈W ⊥ ⇔ ⟨u, w i

⟩= 0 ∀i = 1,2,3...n.

Illustration 8.7 Let R4 have the Euclidean inner product and let u = (−1,1,0,2). Determine whether thevector u is orthogonal to the subspace spanned by the vectors w1 = (1,0,0,0) , w2 = (1,−1,3,0) and w2 =(4,0,9,2) or not ? (or Is u ∈W ⊥ ?)

Solution: In order to check whether given vector is orthogonal to subspce or not, it is sufficient to checkthe orthogonality with each of the spanning vectors [See property 5 of section 8.7].

Since⟨

u, w1⟩= u·w1 = (−1,1,0,2)·(1,0,0,0) = 1 6= 0, u is not orthogonal to w1. Hence u is not orthogonal

to given subspace W. That is u ∉W ⊥.

8.8 Results

Let A be an (m ×n) matrix then

1. The null space of A and the row space of A are orthogonal complements in Rn with respect to Eu-clidean inner product. That is,

W = row(A) ⇔ W ⊥ = null (A)

2. The nullspace of AT and the column space of A are orthogonal complements in Rn with respect toEuclidean inner product. That is

W = col(A) ⇔ W ⊥ = null(

AT )Illustration 8.8 Find the orthogonal complement of subspace ofR3 spanned by the victors (1,−1,3) , (5,−4,−4) , (7,−6,2).

Solution: Given that W = span(1,−1,3) , (5,−4,−4) , (7,−6,2) .

Consider the matrix A by putting given vectors in row, that is A = 1 −1 3

5 −4 −47 −6 2

. Therefore, given sub-

space is W = row(A) and hence its orthogonal complement is W ⊥ = nul(A) [See Result 1 of section 8.8].Now, for null space of A, reducing A to row echelon form.

A = 1 −1 3

5 −4 −47 −6 2

→ R2 −5R1;R3 −7R1

∼ 1 −1 3

0 1 −190 1 −19

→ R3 −R2

∼

1 −1 30 1 −190 0 0

⇒ x = 16t , y = 19t , z = t , t ∈ R

Hence, W ⊥ = nul(A) =

X = (x, y, z

): AX = 0

∴ W ⊥ = (16t ,19t , t ) : t ∈R = span(16,19,1)

[Straight line

]LAVC (GTU-2110015) B.E. Semester II




* Important:

Observe that,

1. Subspace W is the row space of A and hence dimension of W is number of pivot rows of echelon form.Therefore dim(W ) = 2.

2. Orthogonal complement W ⊥ is the null space of A and hence dimension of W ⊥ is number of nonpivot columns of echelon form. That is dim

(W ⊥)= 1.

3. The whole space V is R3 and hence dimension of V is 3. That is dim(V ) = 3.

Thus, dim(W )+dim(W ⊥)= dim(V )

Hence dimension theorem for orthogonal complement is verified.

Illustration 8.9 If subspace W is the intersection of two planes x+ y +z = 0 and x− y +z = 0 in R3, find itsorthogonal complement W ⊥.

Solution: Given W = intersection of two planes x + y + z = 0 and x − y + z = 0 in R3. This can be obtain bysomving system of two eaytions as

x + y + z = 0

x − y + z = 0

⇒ x =−z = t , y = 0, t ∈R

Hence, W = (x, y, z

): x =−z = t , y = 0, t ∈R [

Straight line]

= (t ,0,−t ) : t ∈R = t

∴ W = span(1,0,1) = row(A) where A = [1 0 1

]⇒ W ⊥ = nul(A)

Now for null space of A, matrix A is already in echelon form (because it has only one row). Hence itscorresponding homogeneous system has two parametric solution as x =−t2, y = t1, z = t2, t1, t2 ∈R.Hence, W ⊥ = nul(A)

= (−t2, t1, t2) : t1, t2 ∈R

= t1 (0,1,0)+ t2 (−1,0,1) : t1, t2 ∈R

∴ W ⊥ = span(0,1,0) , (−1,0,1) [Plane]

8.9 Orthogonal Set

â A subset of an inner product space V is called an orthogonal set if all vectors are pairwise orthogonal.That is all pairs of distinct vectors in the set are orthogonal.

Hence, ifu1,u2,u3, ...un

is orthogonal set then,⟨

ui ,u j⟩= 0 ∀i 6= j , and

⟨ui ,ui

⟩ 6= 0

â An orthogonal set in which each vector has unit norm (unit vector) is called an orthonormal set.

e. g. (0,1,0) , (1,0,1) , (1,0,−1) is an orthogonal subset of R3 where as (1,0,0) , (0,1,0) , (0,0,1) is anorthonormal subset of R3.

â On normalizing each vectors of an orthogonal set we get an orthonormal set.

Note: Every orthogonal set (orthonomal set) is always linearly independent and hence, in particular anorthogonal subset of 3 vectors of R3 is always basis for R3. In general it is true for Rn .e. g. (0,1,0) , (1,0,1) , (1,0,−1) is an orthogonal subset of R3 and hence it is basis for R3.





8.10 Orthogonal Projection

Let u and v be vectors of an inner product space V then the orthogo-nal projection of u on v is defined as

pr o jv u =⟨

u, v⟩∥∥v

∥∥2 v

Similarly, orthogonal projection of v on u is defined as

pr o ju v =⟨

v ,u⟩∥∥u

∥∥2 u

Illustration 8.10 Find orthogonal projection of u = (1,−1,2) on v = (2,0,2) with respect to standard Eu-clidean inner product in R3.

Solution: By definition of orthogonal projection,

projv u =⟨

u, v⟩∥∥u

∥∥‖v‖v = u · v∥∥u∥∥‖v‖v

[∵ Standard inner product

]=

((1,−1,2) · (2,0,2)p1+1+4

p4+0+4

)(2,0,2)

=(

2+0+4p6p

8

)(2,0,2)

= 6p48

(2,0,2) = 6

4p

3(2,0,2) =

p3(1,0,1) ⇒ projv u =

(p3,0,

p3)

Exercise 8.2

1. Do there exist k and l such that the vectors u = (2,k,6) , v = (l ,5,3) and w = (1,2,3) are mutually or-thogonal with respect to the Euclidean inner product ?

[Hint: Take u · v = v ·w = w ·u = 0.]

2. Let R3 have the Euclidean inner product, and let u = (1,1,−1) and v = (6,7,−15) . If∥∥ku + v

∥∥ = 13,then what is k ?

[Hint:∥∥ku + v

∥∥2 = 169 ⇒ ⟨ku + v ,ku + v

⟩= 169]

3. Show that p = 1−x +2x2 and q = 2x +x2 are orthogonal in P2.

[Hint: For orthogonal polynomial⟨

p, q⟩= 0.]

4. LetR4 have the Euclidean inner product. Find two unit vectors that are orthogonal to the three vectorsu = (2,1,−4,0) , v = (−1,−1,2,2) and w = (3,2,5,4) .

5. If w is orthogonal to both u1 and u2 then prove that it is orthogonal to k1 u1 +k2 u2, ∀k1,k2 ∈R.

[Hint: Given⟨

w ,u1⟩= ⟨

w ,u2⟩= 0. Prove

⟨w ,k1u1 +k2u2

⟩= 0.]

6. Verify that the set (1,0) , (0,1) is orthogonal with respect to the inner product⟨

u, v⟩= 4u1v1 +u2v2 ,

then convert it to an orthonormal set by normalizing the vectors.

7. Find orthogonal complement ( W ⊥ ) and hence basis for W ⊥. Also verify that dimW +dimW ⊥ =dimV , given

a. W = span(1,4,−2) , (2,1,−1) in R3. b. W = span(1,−1,0,2) , (0,1,2,−1) in R4.

c. W = span(1,−2,1) in R3.

[Hint: See Illustration 8.8]





8. Find the equation of W ⊥, for the each of the following subspace: [Hint: See Illustration 8.9]

a. W be the line in R2 with the equation y = 2x.

b. W be the plane in R3 with the equation x −2y +3z = 0.

c. W be the line in R3 with parametric equation x = 2t , y =−5t , z = 4t , t ∈R.

[Hint: See Illustration 8.9]

Answers

E E E

8.11 Orthogonal and Orthonormal Bases

â A basis B = v1, v2, v2.....vn

of an inner product space V is called an orthogonal basis if it is an or-

thogonal set.

â If each vector of orthogonal basis is unit vector then it is called an orthonormal basis of V.

e. g. (1,0,0) , (0,1,0) , (0,0,1) is an orthonormal basis for R3 whereas (0,2,0) , (1,0,−1) , (1,0,1) is anorthogonal basis for R3 and it can be reduce to orthonormal by normalizing each vector.

8.12 Coordinate Relative to Orthonormal Basis

Let S = v1, v2, v3.....vn

be an orthonormal basis for an inner product space V , and u is any vector in V,

thenu = ⟨

u, v1⟩

v1 +⟨

u, v2⟩

v2 + ...+⟨u, vn

⟩vn

Hence the coordinate of u relative to S is given by(u

)S = (⟨

u, v1⟩

,⟨

u, v2⟩

, .....⟨

u, vn⟩)

Illustration 8.11 Verify that the vectors v1 =(−3

5,

4

5,0

), v2 =

(4

5,

3

5,0

), v3 = (0,0,1) form an orthonormal

basis for R3 with respect to the Euclidean inner product and hence express the vector u = (1,−1,2) as linearcombinations of v1, v2 and v3.

Solution: Observe that for given vectors v1, v2, v3,

⟨v1, v2

⟩= v1 · v2 =(−3

5,

4

5,0

)·(

4

5,

3

5,0

)=

(−3

5

)(4

5

)+

(4

5

)(3

5

)+0 =−12

25+ 12

25= 0

⟨v2, v3

⟩= v2 · v3 =(

4

5,

3

5,0

)· (0,0,1) =

(4

5

)(0)+

(3

5

)(0)+ (0)(1) = 0

⟨v3, v1

⟩= v3 · v1 = (0,0,1) ·(−3

5,

4

5,0

)= (0)

(−3

5

)+ (0)

(4

5

)+ (1)(0) = 0

Further,

‖v1‖ =√(

−3

5

)2

+(

4

5

)2

+ (0)2 =√

9

25+ 16

25+0 =

√25

25=p

1 = 1

‖v2‖ =√(

4

5

)2

+(

3

5

)2

+ (0)2 =√

16

25+ 9

25+0 =

√25

25=p

1 = 1

‖v3‖ =√

(0)2 + (0)2 + (1)2 =p0+0+1 =p

1 = 1





∴ Given set is an orthonormal subset of R3 containing three vectors. So it is an orthonarmal basis for R3.

Now for orthonormal basis, we have [See section 8.8]

u = ⟨v1,u

⟩v1 +

⟨v2,u

⟩v2 +

⟨v3,u

⟩v3

= (v1 ·u

)v1 +

(v2 ·u

)v2 +

(v3 ·u

)v3

[∵ Standard Euclidean inner product

]=

[(−3

5,

4

5,0

)· (1,−1,2)

]v1 +

[(4

5,−3

5,0

)· (1,−1,2)

]v2 + [(1,0,0) · (1,−1,2)] v3 [∵ Given]

∴ u =−7

5v1 + 7

5v2 + v3 Required linear combination.

Also coordinate of u relative to given orthonormal basis S = v1, v2, v3

is

(u

)S =

(−7

5,

7

5,1

).

8.13 Gram-Schmidt Process 1

â With the help of this process we can construct an orthogonal basis from the given basis and on nor-malizing each vector we can obtain an orthonormal basis.

â Consider the basis S = u1,u2,u3.....un

of an inner product space V.

â The orthogonal basis of an inner product space V is given by B = w1, w2, w3.....wn

, where

w1 = u1

w2 = u2 −projw 1u2

= u2 −⟨

w1,u2⟩∥∥w1

∥∥2 w1

w3 = u3 −projw 1u3 −projw 2

u3

= u3 −⟨

w1,u3⟩∥∥w1

∥∥2 w1 −⟨

w2,u3⟩∥∥w2

∥∥2 w2 and so on.

Note that on normalizing each vector using w = w∥∥w∥∥ , we get an othonormal basis B = w1, w2, w3, ...wn .

Illustration 8.12 Let R3 have Euclidean inner product. Transform the basis S = u1,u2,u3

into an or-

thonormal basis using Gram-Schmidt process, where u1 = (1,0,0) ,u2 = (3,7,−2) and u3 = (0,4,1) . [Summer-2017]

Solution: Given inner product is standard Euclidean inner product, that is⟨

u, v⟩= u · v .

By Gram-Schmidt process,

w1 = u1 = (1,0,0) ⇒ w1 = w1∥∥w1∥∥ = (1,0,0)p

1+0+0= 1p

1(1,0,0) ∴ w1 = (1,0,0)

w2 = u2 −projw 1u2

= u2 −⟨

w1,u2⟩∥∥w1

∥∥2 w1 = u2 − w1 ·u2∥∥w1∥∥2 w1

[∵

⟨w1,u2

⟩= w1 ·u2]

= (3,7,−2)− (1,0,0) · (3,7,−2)

(1)(1,0,0)

= (3,7,−2)−3(1,0,0)

∴ w2 = (0,7,−2) ⇒ w2 = w2∥∥w2∥∥ = (0,7,−2)p

0+49+4= 1p

53(0,7,−2) ∴ w2 =

(0,

7p53

,− 2p53

)1Jorgen Pedersen Gram; Danish, 1850-1916 and Erhard Schmidt; Berlin, 1876-1959.





w3 = u3 −projw 1u3 −projw 2

u3

= u3 −⟨

w1,u3⟩∥∥w1

∥∥2 w1 −⟨

w2,u3⟩∥∥w2

∥∥2 w2 = u3 − w1 ·u3∥∥w1∥∥2 w1 − w2 ·u3∥∥w2

∥∥2 w2

= (0,4,1)− (1,0,0) · (0,4,1)

(1)(1,0,0)− (0,7,−2) · (0,4,1)

53(0,7,−2)

= (0,4,1)−0− 26

53(0,7,−2) = (0,4,1)−

(0,

182

53,−52

53

)w3 =

(0,

30

53,−105

53

)= 15

53(0,2,7)

⇒ w3 = w3∥∥w3∥∥ = (0,2,7)p

0+4+49= 1p

53(0,2,7) ∴ w3 =

(0,

2p53

,7p53

)

∴ Required orthonormal basis is B = w1, w2, w3 =

(1,0,0) ,

(0,

7p53

,− 2p53

),

(0,

2p53

,7p53

).

Note: In case of orthogonal basis we need not to find normalized vector. Hence orthogonal basis is B =w1, w2, w3

= (1,0,0) , (0,7,−2) ,

(0,

30

53,−105

53

).

Illustration 8.13 LetR3 have an Euclidean inner product, Find the orthonormal basis for the space spannedby (0,1,2) , (−1,0,1) , (−1,1,3) .

Solution: Let W = span(0,1,2) , (−1,0,1) , (−1,1,3)In order to find orthogonal basis of W, first of all we find basis for W.

Observe that, for the matrix of column vectors of given set A = 0 −1 −1

1 0 12 1 3

, det(A) = 0. therefore

given set is linearly dependent and hence it is not a basis. Now to remove linearly dependent vector reducingmatrix A to row echelon form, we get

A = 0 −1 −1

1 0 12 1 3

∼

1 0 10 −1 −10 0 0

Discarding third vector corresponding to non pivot column from given set we get basis for subspace W asS = (0,1,2) , (−1,0,1) =

u1,u2

.

Now by Gram-Schmidt method,

w1 = u1 = (0,1,2) ⇒ w1 = w1∥∥w1∥∥ = 1p

5(0,1,2) ∴ w1 =

(0,

1p5

,2p5

)

w2 = u2 −projw 1u2 = u2 −

⟨w1,u2

⟩∥∥w1∥∥2 w1 = u2 − w1 ·u2∥∥w1

∥∥2 w1[∵ Eulidean inner product

]= (−1,0,1)− (0,1,2) · (−1,0,1)(

0+12 +22) (0,1,2) = (−1,0,1)− 2

5(0,1,2)

= (−1,0,1)−(0,

2

5,

4

5

)=

(−1,−2

5,

1

5

)w2 = 1

5(−5,−2,1) ⇒ w2 = w2∥∥w2

∥∥ = 1p30

(−5,−2,1) ∴ w2 =( −5p

30,−2p

30,

1p30

)

∴ Required orthonormal basis for given subspace W is B = w1, w2 =(

0,1p5

,2p5

),

( −5p30

,−2p

30,

1p30

).





Exercise 8.3

1. In each part, an orthonormal basis relative to the Euclidean inner product is given, find coordinate ofw with respect to that basis:

a. w = (3,7) , u1 =(

1p2

,− 1p2

),u2 =

(1p2

,1p2

).

b. w = (−1,0,2) , u1 =(

2

3,−2

3,

1

3

),u2 =

(2

3,

1

3,−2

3

),u3 =

(1

3,

2

3,

2

3

).

[See Illustration 8.11]

2. Use Gram-Schmidt process to transform the given basis in to an orthonormal basis:

a. u1 = (1,−3) ,u2 = (2,2) in R2, with Euclidean inner product.

b. u1 = (1,1,1) ,u2 = (−1,1,0) ,u3 = (1,2,1) in R3, with Euclidean inner product.


3. LetR3 have an Euclidean inner product, Find the orthonormal basis for the space spanned by (1,−1,2) ,(1,1,0) , (1,0,1) .


4. Let

v1, v2, v3

be the orthonormal basis for an inner product space V . Show that∥∥w

∥∥2 = ⟨w , v1

⟩2 +⟨w , v2

⟩2 +⟨w , v3

⟩2,∀w ∈V.

[Hint: For orthonormal basis S,(w

)S = (⟨w , v1⟩ ,⟨w , v2⟩ ,⟨w , v3⟩) . ]

Answers

1. a.(−2

p2,5

p2)

b.(0,−4

3,1

)

2. a. B =(

1p10

,− 3p10

),

(3p10

,1p10

)b. B =

(1p3

,1p3

,1p3

),

(− 1p

2,

1p2

,0

),

(1p6

,1p6

,− 2p6

)

3. B =(

1p6

,− 1p6

,2p6

),

(1p2

,1p2

,0

)E E E

Theorem 8.1 (Projection Theorem | Orthogonal Projection on a Subspace)Let W be a finite dimensional subspace of an inner product space V with an orthonormal basis

v1, v2, v3, ...vn

.

Then every vector u ∈V can be uniquely expressed as

u = w1 +w2

where,

â w1 is called orthogonal projection of u on W and is denote by projW u and is given by

w1 = projW u = ⟨u, v1⟩ v1 +⟨u, v2⟩ v2 +⟨u, v3⟩ v3 + ....+⟨u, vn⟩ vn

â w2 is called component of u orthogonal to W and is denote by pr o jW ⊥u. that is w2 = pr o jW ⊥u ∈W ⊥.Thus

u = projW u +projW ⊥ u

Note: In case of orthogonal basis first transform in to orthonormal basis by normalizing each vector andthen proceed further.





Projection on lineProjection on plane

Illustration 8.14 Let W be subspaceR3 spanned by the orthogonal vectors v1 = (0,1,0) and v2 =(−4

5,0,

3

5

).

Find the projection of u = (1,1,1) on W. also obtain the component of u orthogonal to W.

Solution: For given orthogonal vectors v1 = (0,1,0) and v2 =(−4

5,0,

3

5

), observe that ‖v1‖ = ‖v2‖ = 1.

Hence v1, v2 form an orthonormal basis for subspace W.∴ Orthogonal projection of u = (1,1,1) on W is given by

projW u = ⟨u, v1⟩ v1 +⟨u, v2⟩ v2 [∵Theorem 8.1]

= (v1 · u) v1 + (v2 · u) v2[∵ Euclidean inner product

]= [(1,1,1) · (0,1,0)] (0,1,0)+

[(1,1,1) ·

(−4

5,0,

3

5

)](−4

5,0,

3

5

)= [1](0,1,0)+

[−1

5

](−4

5,0,

3

5

)= (0,1,0)+

(4

25,0,− 3

25

)∴ projW u =

(4

25,1,− 3

25

)Also by Projection Theorem 8.1, we have

u = projW u +projW ⊥u

∴ projW ⊥u = u −projW u

= (1,1,1)−(

4

25,1,− 3

25

)=

(1− 4

25,0,1+ 3

25

)∴ projW ⊥u =

(21

25,0,

28

25

)Required component of u orthogonal to W.

Exercise 8.4

1. The subspace of R3 spanned by the vectors u1 =(4

5 ,0,−35

)and u2 = (0,1,0) is a plane passing through

the origin. Express w = (1,2,3) in the form w = w1 +w2, where w1 lies in the plane and w2 is perpen-dicular to the plane.

2. Let W be The subspace of R4 spanned by the vectors u1 = (−1,0,1,0) and u2 = (0,1,0,1), Express w =(−1,2,6,0) in the form w = w1 +w2, where w1 lies in subspace W and w2 is orthogonal to W.

Answers

1. w1 =(−4

5,2,

3

5

), w2 =

(9

5,0,

12

5

)2. w1 =

(−7

2,1,

7

2,1

), w2 =

(5

2,1,

5

2,−1

)E E E





8.14 Least Square Approximate Solution for Linear System

â Let AX = b be the inconsistent system of linear equations. That is its exact solution does not exist.

â The beast approximate solution of AX = b is known as the least square solution and is given by thenormal system,

AT AX = AT b (8.3)

â Also if W denotes the column space of A and X be the least square solution then the orthogonalprojection of b on W is given by

projW b = AX (8.4)

Illustration 8.15 Find the least squares solution of the linear system Ax = b for A = 4 0

0 21 1

,b = 2

011

.

Also find projection of b on column space of A. [Winter-2015]

Solution: The least square solution is given by (8.3),

AT AX = AT b,(

X ∈R2)

[4 0 10 2 1

] 4 00 21 1

[xy

]=

[4 0 10 2 1

] 20

11

[

17 11 5

][xy

]=

[1911

]⇒ 17x + y = 19, x +5y = 11

Solving above two equations, we get required least square solution as x = 1, y = 5.Also projection of b on column space of A is given by (8.4),

projW b = AX , where W = col(A) and X is the least square solution.

∴ projW b = 4 0

0 21 1

[15

]=

4106

∴ projW b = (4,10,6)

Illustration 8.16 Find the orthogonal projection of the vector u = (−3,−3,8,9) on the subspace of R4

spanned by the vectors u1 = (3,1,0,1) ,u2 = (1,2,1,1) ,u3 = (−1,0,2,−1) .

Solution: Let W = span u1, u2, u3 = span(3,1,0,1) , (1,2,1,1) , (−1,0,2,−1) .

Then W = col(A) , where A be matrix obtained by putting vectors in columns, that is A =

3 1 −11 2 00 1 21 1 −1

.

∴ By least square method, required orthogonal projection is given by

projW u = AX (8.5)

where X is the least square solution of the system

AX = u ⇒

3 1 −11 2 00 1 21 1 −1

x

yz

=

−3−389





The normal system for the least square solution:

AT AX = AT u 3 1 0 11 2 1 1

−1 0 2 −1

3 1 −11 2 00 1 21 1 −1

x

yz

= 3 1 0 1

1 2 1 1−1 0 2 −1

−3−389

11 6 −4

6 7 0−4 0 6

xyz

= −3

810

(8.6)

To solve system (8.6), reducing corresponding augmented matrix to row echelon form we get

[A : u

]= 11 6 −4 −3

6 7 0 8−4 0 6 10

∼ 11 6 −4 −3

0 41 24 1060 0 1 1

Making back substitution, we get the least square solution as x =−1, y = 2, z = 1.∴ From (8.5), required orthogonal projection is

projW u =

3 1 −11 2 00 1 21 1 −1

−1

21

=

−2

340

∴ projW u = (−2,3,4,0)

Exercise 8.5

1. Find the least squares solution of the linear system Ax = b for A = 2 −2

1 13 1

,b = 2

−11

. Also find

projection of b on column space of A. [Winter-2015]

2. Find the least square solution for the system 4x1−3x2 = 12, 2x1+5x2 = 32, 3x1+x2 = 21. [Summer-2015]

3. Find the orthogonal projection of the vector u = (6,3,9,6) onto the subspace of R4 spanned by thevectors u1 = (2,1,1,1) ,u2 = (1,0,1,1) ,u3 = (−2,−1,0,−1) .

4. Find projW u, where u = (5,6,7,2) and W is the solution space of the homogeneous system x1+x2+x3 =0, 2x2 +x3 +x4 = 0.

[Hint: Solution space: W = span

(−1

2,−1

2,1,0

),

(1

2,−1

2,0,1

)]

Answers

1. x = 3

7, y =−2

3,projcol(A)b =

(46

21,− 5

21,

13

21

)2. x = 305

39, y = 704

2733. projW u = (7,2,9,5)

4. projW u = (0,−1,1,1)

E E E

8.15 Orthogonal Matrix

A square matrix A is said to be orthogonal matrix if, A−1 = AT , that is A AT = AT A = I .e. g.

A =

1p2

1p2

− 1p2

1p2

, A = cosθ −sinθ

sinθ cosθ

are orthogonal matrices.





* Important:

A matrix of order n is an orthogonal matrix if and only if its row (column) vectors form orthonormal subsetof Rn .e. g. Consider orthonormal subset

u1,u2,u3

of R3 where u1 =

(1p2

,0, 1p2

),u2 = (0,1,0) ,u3 =

(− 1p

2,0, 1p

2

),

then A =

1p2

0 1p2

0 1 0

− 1p2

0 1p2

and AT =

1p2

0 − 1p2

0 1 0

1p2

0 1p2

are always orthogonal matrices.

Note: In order to check weather given matrix is orthogonal or not it is sufficient to check orthonoamal-ity of its row or column vectors.

Illustration 8.17 Show that the matrix A =

13

23

23

23 −2

313

−23 −1

323

is orthogonal matrix and hence find A−1.

Solution: Consider the column vectors u1 =(

1

3,

2

3,−2

3

), u2 =

(2

3,−2

3,−1

3

), u3 =

(2

3,

1

3,

2

3

).

Observe that, u1 ·u2 =(

1

3,

2

3,−2

3

)·(

2

3,−2

3,−1

3

)= 2

9− 4

9+ 2

9= 0

u2 ·u3 =(

2

3,−2

3,−1

3

)·(

2

3,

1

3,

2

3

)= 4

9− 2

9− 2

9= 0

u1 ·u3 =(

1

3,

2

3,−2

3

)·(

2

3,

1

3,

2

3

)= 2

9+ 2

9− 4

9= 0

Also,∥∥u1

∥∥=√(

1

3

)2

+(

2

3

)2

+(−2

3

)2

=√

1

9+ 4

9+ 4

9= 1

∥∥u2∥∥=

√(2

3

)2

+(−2

3

)2

+(−1

3

)2

=√

4

9+ 4

9+ 1

9= 1

∥∥u3∥∥=

√(2

3

)2

+(

1

3

)2

+(

2

3

)2

=√

4

9+ 1

9+ 4

9= 1

∴u1,u2,u3

forms orthonormal subset of R3. Hence A is an orthogonal matrix, and

A−1 = AT =

13

23 −2

3

23 −2

3 −13

23

13

23

Illustration 8.18 Is A =

2 2 1

−2 1 2

1 −2 2

orthogonal matrix ? if not can it be converted in to orthogonal

matrix ? [Summer-2015]

Solution: Consider the column vectors u1 = (2,−2,1) , u2 = (2,1,−2) , u3 = (1,2,2) . Observe that u1 ·u2 = u2 ·u3 = u1 ·u3 = 0 and

∥∥u1∥∥ = ∥∥u2

∥∥ = ∥∥u3∥∥ = 3. That is column vectors are orthogonal but not or-

thonormal (not unit vector). Hence, A is not orthogonal matrix.





Since column vectors are orthogonal, on normalizing each vector by dividing its magnitude, we obtainorthonormal vectors. Thus given matrix A can be converted into orthogonal matrix, and is given by

2/3 2/3 1/3

−2/3 1/3 2/3

1/3 −2/3 2/3

Illustration 8.19 If A is an orthogonal then prove that det(A) = ±1. Also show that converse may not betrue.

Solution: Suppose that A is an orthogonal matrix.Therefore, AT = A−1

∴ A AT = I ⇒ det(

A AT )= det(I )

∴ det(A)det(A) = 1[∵ det(I ) = 1 as I is an identity matrix

]∴ [det(A)]2 = 1 ⇒ det(A) =±1 Proved.

Now consider the matrix A = 2 1

1 1

. Here det(A) = 1 but matrix is not orthogonal because its column

vectors are not orthonormal. Thus converse of given statement may not be true.

Exercise 8.6

1. Show that the matrix A is orthogonal then AT and A−1 are also orthogonal.

[Hint: Apply definition of orthogonal matrix.]

2. Find the normal system of Ax = b when A is orthogonal matrix.

3. Let A be a square matrix such that A2 = I . Prove that A is symmetric if and only if A is orthogonal.

4. Let A is an (n ×n) orthogonal matrix with n is odd, then prove that A cannot be skew-symmetric.

5. Let A is an (n ×n) orthogonal matrix such that| A | = −1. Prove that (A+ In) is singular matrix.

E E E

Powered by





Contact: (+91) 9427 80 9779





Chapter 9Vector calculus I: Vector differentiation

9.1 Scalar and Vector

A physical quantity for the representation of it only magnitude is sufficient is called scalar whereas a physicalquantity for the representation of it magnitude as well as direction is required, is called vector.e. g. statistical data are scalars and velocity, acceleration, force etc are vectors.

General Remarks

1. A scalar is generally denoted by α,β, a,b etc. whereas the vector is denoted by−→A , A, A etc.

2. A vector having unit magnitude is called unit vector and is denoted by a or A (read as cap or carat).In particular the unit vectors along the positive direction of the coordinate axes, X -axis, Y -axis andZ-axis, are denoted by i , j , k or I , J , K respectively.

3. Any vector A can always be expressed as a combination of i , j , k, that is−→A = a1 i +a2 j +a3k for some

a1, a2, a3 ∈ R. This form of vector is known as component form and a1, a2, a3 are called componentalong respective coordinate axes.

4. The magnitude of a vector−→A is a scalar and denoted by the symbol or A and is given by the formula∣∣∣−→A ∣∣∣= A =+

√a2

1 +a22 +a2

3. (only positive value)

5. Dividing the vector by its own magnitude we get a unit vector along the direction of given vector, that

is unit vector along the direction of−→A is given by

a = A =−→A

A= a1 i +a2 j +a3k√

a21 +a2

2 +a23

e. g.−→A = 2i − j +3k ⇒ a = 1p

13

(2i − j +3k

)9.2 Algebraic Operations of Vectors

For−→A = a1 i +a2 j +a3k,

−→B = b1 i +b2 j +b3k and α ∈R, the basic algebraic operations are defined as follow:

1. Scalar Multiplication:

α−→A =α(

a1 i +a2 j +a3k)= (αa1) i + (αa2) j + (αa3) k

2. Vector Addition and subtraction:

−→A ±−→

B = (a1 i +a2 j +a3k

)± (b1 i +b2 j +b3k

)= (a1 i +a2 j +a3k

)± (b1 i +b2 j +b3k

)104




3. Vector Multiplications If θ denotes an angle between the directions of two vectors−→A and

−→B then

there defined two types of product between−→A and

−→B as

i. Dot Product (Scalar Product):−→A ·−→B = A B cosθ = a1b1 +a2b2 +a3b3

ii. Cross Product (Vector Product):

−→A ×−→

B = A B sinθ n =

∣∣∣∣∣∣∣∣∣∣i j k

a1 a2 a3

b1 b2 b3

∣∣∣∣∣∣∣∣∣∣where n is the unit vector perpendicular to both

−→A and

−→B and in direction in which the right

handed screw would advance when it rotate from−→A to

−→B .

9.3 Point Functions

â A function whose value depends on the position of the point in space is called point function.

â If that function is scalar then it is called scalar point function.

â If that function is vector then the function is called vector point function.e. g. Temperature in the medium is scalar point function and the velocity of particle in the movingfluid is the vector point function.

â In symbolic form, φ(x, y, z

) = x y2 +7x y z3 is scalar point function and−→V = 3xzi −5x y2 j + x y z3k is

the vector point function defined at the point p(x, y, z

).

9.4 Vector Differential Operator

An operator of the form∂

∂xi + ∂

∂yj + ∂

∂zk is called the vector differential operator and is denoted by the

symbol ∇ (read as del or nabla). That is,

∇= ∂

∂xi + ∂

∂yj + ∂

∂zk

9.5 Gradient

Let φ(x, y, z

)be the scalar point function then the gradient of φ

(x, y, z

)is defined as gradφ=∇φ.

∴ gradφ=∇φ=(∂

∂xi + ∂

∂yj + ∂

∂zk

)φ= ∂φ

∂xi + ∂φ

∂yj + ∂φ

∂zk

Observe that gradient is defined for the scalar point function and it gives vector point function.

Geometrical Interpretation of Gradient

If p(x, y, z

)be any point on the given surface φ

(x, y, z

) = c then(∇φ)P defines normal vector to the surface, and hence the unit

normal vector to the surface φ= c at the point p is given by,

n =(∇φ)

p∣∣∣(∇φ)p

∣∣∣





Illustration 9.1 Find gradφ if φ= log(x2 + y2 + z2) at the point (1,2,1) .

Solution: By definition of gradient vector,

gradφ=∇φ

= ∂φ

∂xi + ∂φ

∂yj + ∂φ

∂zk

=(

2x

x2 + y2 + z2

)i +

(2y

x2 + y2 + z2

)j +

(2z

x2 + y2 + z2

)k

[∵ φ= log

(x2 + y2 + z2)]

∴ gradφ= 2

x2 + y2 + z2

(xi + y j + zk

)At

(x, y, z

)= (1,2,1) ,

gradφ= 2

6

(i +2 j + k

)= 1

3i + 2

3j + 1

3

Illustration 9.2

If r = ∣∣−→r ∣∣ , where −→r = xi + y j + zk, prove that∇ f (r ) = f ′ (r )∇r. Hence deduce that ∇∣∣−→r ∣∣2 = 2~r .

Solution: Given~r = xi + y j + zk and r = ∣∣−→r ∣∣=√x2 + y2 + z2.

∇ f (r ) =(∂

∂xi + ∂

∂yj + ∂

∂zk

)f (r )

= ∂

∂xf (r ) i + ∂

∂yf (r ) j + ∂

∂zf (r ) k

= f ′ (r )∂r

∂xi + f ′ (r )

∂r

∂yj + f ′ (r )

∂r

∂zk

[∵ By chain rule for partial derivative

]= f ′ (r )

(∂r

∂xi + ∂r

∂yj + ∂r

∂zk

)(9.1)

= f ′ (r )

(∂

∂xi + ∂

∂yj + ∂

∂zk

)r

∴ ∇ f (r ) = f ′ (r )∇r

[∵

(∂

∂xi + ∂

∂yj + ∂

∂zk

)=∇

]

Also, r =√

x2 + y2 + z2 ⇒ ∂r

∂x= 2x

2√

x2 + y2 + z2= x

r. Similarly,

∂r

∂y= y

rand

∂r

∂z= z

r.

Substituting in (9.1),

∇ f (r ) = f ′ (r )( x

r

)i + f ′ (r )

( y

r

)j + f ′ (r )

( z

r

)k

= f ′ (r )

r

(xi + y j + zk

)= f ′ (r )

r(~r )

∴ ∇ f (r ) = f ′ (r )~r

r(9.2)

Put f (r ) = ∣∣−→r ∣∣2 = r 2 ⇒ f ′ (r ) = 2r ∴ ∇∣∣−→r ∣∣2 = 2r~r

r= 2~r

Illustration 9.3 Find the unit normal vector to the surface x y3z2 = 4 at the point (−1,−1,2) .

Solution: Let φ = x y3z2 − 4 (Taking all terms of given surface x y3z2 = 4 on one side) and given pointp (−1,−1,2) .Unit normal vector to given surface at a point p is [See section 9.5]

n =(∇φ)

p∣∣∣(∇φ)p

∣∣∣ (9.3)





Now,

∇φ= ∂φ

∂xi + ∂φ

∂yj + ∂φ

∂zk

= (y3z2) i + (

3x y2z2) j + (2x y3z

)k

[∵φ= x y3z2 −4

]At the point p (−1,−1,2) ,(∇φ)

p =−4i −12 j +4k = 4(−i −3 j + k

)(9.4)

∴∣∣∇φ∣∣

p =√

(−4)2 + (−12)2 + (4)2 = 4p

11 (9.5)

Substituting the values from (9.5) and (9.4) in (9.3), we get required unit normal vector,

n = 4(−i −3 j + k

)4p

11∴ n = 1p

11

(−i −3 j + k)

9.6 Divergence

Let−→V = V1 i +V2 j +V3k be the vector point function then the divergence of

−→V

(x, y, z

)is defined as div~V =

∇·~V .

∴ div~V =∇·~V =(∂

∂xi + ∂

∂yj + ∂

∂zk

)· (V1 i + v2 j +V3k

)= ∂V1

∂x+ ∂V2

∂y+ ∂V3

∂z

Observe that divergence is defined for the vector point function and it gives scalar.

Physical Interpretation of Divergence

â If−→V denotes the linear velocity of the particle of moving fluid then div~V =∇·~V defines the rate of

increase of fluid per unit volume at a point p.

Thus, we can say that div~V =∇·~V gives the rate at which the fluid is originating (diverges) from thepoint per unit volume.

â If the divergence of velocity is zero i.e. div~V =∇·~V = 0, then such a fluid is known as solenoidal orincompressible.

Illustration 9.4 Evaluate div(3x2 i +5x y2 j +x y z3k

)at (1,2,3) .

Solution: Let−→V = (

3x2 i +5x y2 j +x y z3k)=V1 i +V2 j +V3k

∴ div−→V = ∂V1

∂x+ ∂V2

∂y+ ∂V3

∂z

[∵ By definition

]= (6x)+ (

10x y)+ (

3x y z2)At the point (1,2,3) , div

−→V = 6(1)+10(2)+3(18) = 80.

Illustration 9.5 Find the value ofα such that the vector−→V = (

αx2 y + y z)

i+(x y2 −xz2) j+(

2x y z −2x2 y2) kis solenoidal.

Solution: We know that a vector−→V is solenoidal (incompressible) if div

−→V = 0.

∴ div((αx2 y + y z

)i + (

x y2 −xz2) j + (2x y z −2x2 y2) k

)= 0

∴∂

∂x

(αx2 y + y z

)+ ∂

∂y

(x y2 −xz2)+ ∂

∂z

(2x y z −2x2 y2)= 0

∴(2αx y +0

)+ (2x y −0

)+ (2x y −0

)= 0

∴ 2αx y +4x y = 0 ⇒ 2α+4 = 0 ∴ α=−2





9.7 Curl

Let−→V =V1 i +V2 j +V3k be the vector point function then the curl of

−→V

(x, y, z

)is defined as curl~V =∇×~V .

∴ curl~V =∇×~V =(∂

∂xi + ∂

∂yj + ∂

∂zk

)× (

V1 i + v2 j +V3k)=

∣∣∣∣∣∣∣∣∣∣∣i j k∂

∂x

∂

∂y

∂

∂z

V1 V2 V3

∣∣∣∣∣∣∣∣∣∣∣Observe that curl is defined for the vector point function and it gives again vector.

Physical Interpretation of Curl

Let p(x, y, z

)be any particle of the rotating rigid body, about some axis

L. Suppose that−→V be its linear velocity and

−→Ω be its angular velocity

then−→Ω = 1

2Cul r

−→V

Thus, we can say that for the rotating body the angular velocity is thehalf curl of its linear velocity.

â If the angular velocity of the rotating body is zero then there no rotation.

∴ We have,−→Ω =−→

0 ⇒ Culr−→V =−→

0 .

â Such a motion is known as irrotational motion and vector filed is known as irrotationa field.

â For an irrotational field always there exist a scalar function (scalar potential function) φ such that

−→V = gradφ=∇φ

â Such a system is called conservative system, that is work does not depend on the path.

Illustration 9.6 Find curl−→F , if

−→F = (

y2 cos x + z2) i + (2y sin x −4

)j +3xz2k. Is

−→F irrotational ? [Summer-

2016]

Solution: Let−→F = (

y2 cos x + z2) i + (2y sin x −4

)j +3xz2k = F1 i +F2 j +F3k.

By definition of curl,

curl−→F =∇×−→

F

=

∣∣∣∣∣∣∣∣∣∣∣i j k∂

∂x

∂

∂y

∂

∂z

F1 F2 F3

∣∣∣∣∣∣∣∣∣∣∣=

∣∣∣∣∣∣∣∣∣∣∣i j k∂

∂x

∂

∂y

∂

∂z(y2 cos x + z2) (

2y sin x −4)

3xz2

∣∣∣∣∣∣∣∣∣∣∣= i

[∂

∂y

(3xz2)− ∂

∂z

(2y sin x −4

)]− j

[∂

∂x

(3xz2)− ∂

∂z

(y2 cos x + z2)]





+ k

[∂

∂x

(2y sin x −4

)− ∂

∂y

(y2 cos x + z2)]

= i [0−0]− j[(

3z2)− (2z)]+ k

[(2y cos x

)− (2y cos x

)]∴ curl

−→F = (

2z −3z2) j

Since curl−→F 6= −→

0 ,−→F is not an irrotational vector field.

Illustration 9.7 Show that the vector field−→A = (

x2 − y2 +x)

i − (2x y + y

)j is irrotational. Also find scalar

function φ such that−→A = gradφ.

Solution: Given−→A = (

x2 − y2 +x)

i − (2x y + y

)j .

curl−→A =∇×−→

A

=

∣∣∣∣∣∣∣∣∣∣∣i j k∂

∂x

∂

∂y

∂

∂z(x2 − y2 +x

) (−2x y −2y)

0

∣∣∣∣∣∣∣∣∣∣∣[∵ Coefficient of k is 0

]

= i

[∂

∂y(0)− ∂

∂z

(−2x y −2y)]− j

[∂

∂x(0)− ∂

∂z

(x2 − y2 +x

)]+ k

[∂

∂x

(−2x y −2y)− ∂

∂y

(x2 − y2 +x

)]= i [0−0]− j [0−0]+ k

[(−2y)− (−2)

]= (0) i − (0) j + (0) k =−→0

∴ curl−→A =−→

0 ⇒ −→A is an irrotational.

Since−→A is an irrotational field, there exist a scalar function (called scalar potential function) φ such that

−→A = gradφ=∇φ

= ∂φ

∂xi + ∂φ

∂yj + ∂φ

∂zk

[∵ Definition of gradient

]Equating with components of

−→A , we get

∂φ

∂x= x2 − y2 +x,

∂φ

∂y=−2x y −2y,

∂φ

∂z= 0

Integrating above equations partially with respect to x, y, z respectively, keeping other variable constant, weget following equations:

φ= x3

3−x y2 + x2

2+ c1

(y, z

) [From

∂φ

∂xkepping y, z constant

](9.6)

φ=−x y2 − y2 + c2 (x, z)

[From

∂φ

∂ykepping x, z constant

](9.7)

φ= c3(x, y

) [From

∂φ

∂zkepping x, y constant

](9.8)

in (9.6), c1(y, z

)be the terms of φ not containing x, and be taken from (9.7) and (9.8), that is c1

(y, z

)=−y2.

Hence from (9.6) required scalar function for which−→A = gradφ, is

φ= x3

3−x y2 + x2

2− y2

Note: Instead of c1(y, z

), if we find c2 (x, z) or c3

(x, y

)from other two equations using same logic, we will

get the same answer. More precisely, just add all terms of φ (without c1,c2,c3) by taking each term exactlyonce. (Verify !)





9.8 Directional Derivative

Let f(x, y, z

)be the scalar point function and

−−→PQ be the given direc-

tion then the directional derivative of f(x, y, z

)at the point p in the

direction of−−→PQ is denoted by

∂ f

∂rand is defined as

∂ f

∂r= (∇ f

)P · N (9.9)

whereN is unit vector along the given direction−−→PQ, that is N =

−−→PQ∣∣∣−−→PQ

∣∣∣Also if θ denotes the angle between

(∇ f)

P and N then from (9.9), we have

∂ f

∂r= ∣∣(∇ f

)P

∣∣ ∣∣N ∣∣cosθ = ∣∣∇ f∣∣P cosθ

[∵ N is unit vector, so

∣∣N ∣∣= 1]

⇒ max∂ f

∂r= ∣∣∇ f

∣∣P when cosθ = 1 (i.e. θ = 0)

Thus, we conclude that the maximum directional derivative (the rate of change) of f(x, y, z

)occurs along

the direction of(∇ f

)P and its magnitude is equals to

∣∣∇ f∣∣P .

Illustration 9.8 Find the directional derivative of the function f = x y2 + y z3 at the point (2,−1,1) in thedirection of the vector i +2 j +2k. [Summer-2017]

Solution: Given f = x y2 + y z3, P (2,−1,1)−−→PQ = i +2 j +2k

By definition, required directional derivative is

∂ f

∂r= (∇ f

)P · N (9.10)

where, N = Unit vector along given direction−−→PQ = i +2 j +2k

=−−→PQ∣∣∣−−→PQ

∣∣∣ = i +2 j +2kp1+4+4

∴ N = 1

3

(i +2 j +2k

)Also, ∇ f = ∂ f

∂xi + ∂ f

∂yj + ∂ f

∂zk

= ∂

∂x

(x y2 + y z3) i + ∂

∂y

(x y2 + y z3) j + ∂

∂z

(x y2 + y z3) k

= (y2) i + (

2x y + z3) j + (3z2) k

∴(∇ f

)P = i −3 j +3k [∵ P (2,−1,1)]

Substituting the values of(∇ f

)P and N in (9.10),

∂ f

∂r= (

i −3 j +3k) · 1

3

(i +2 j +2k

)= 1

3(1−6+6) = 1

3∴

∂ f

∂r= 1

3

Note: The maximum directional derivative is∣∣∇ f

∣∣P = p

1+9+9 = p19 and it occurs in the direction of

gradient vector(∇ f

)P = i −3 j +3k. Direction may be given by unit vector, that is

(∇ f)

P∣∣∇ f∣∣P

= 1p19

(i −3 j +3k

).





9.9 Angle between two Surfaces

An angle θ between two surfaces (tangent planes) at a given point is defined as an acute angle(0 É θ É π

2

)between two normals at that point. The cosine of θ it is given by

cosθ =−→N 1 ·−→N 2∣∣∣−→N 1

∣∣∣ ∣∣∣−→N 2

∣∣∣ (cosθ Ê 0)

Illustration 9.9 Find the angle between the surfaces x2+y2+z2 = 9 and z = x2+y2−3 at the point (2,−1,2) .

Solution: Let φ1 = x2 + y2 + z2 −9, φ1 = z −x2 − y2 +3, p (2,−1,2)

∇φ1 = ∂φ1

∂xi + ∂φ1

∂yj + ∂φ1

∂zk = 2xi +2y j +2zk

∇φ2 = ∂φ2

∂xi + ∂φ2

∂yj + ∂φ2

∂zk =−2xi −2y j + k

At the point p (2,−1,2) , normal to the given surfaces are

−→N 1 =

(∇φ1)

p = 4i −2 j +4k and−→N 2 =

(∇φ2)

p =−4i +2 j + k

Now, angle between two surfaces is given by

cosθ =−→N 1 ·−→N 2∣∣∣−→N 1

∣∣∣ ∣∣∣−→N 2

∣∣∣ (cosθ Ê 0)

=(4i −2 j +4k

) · (−4i +2 j + k)

p16+4+16

p16+4+1

= −16−4+4

6p

21

∴ cosθ =− 8

3p

21

Since θ is an acute angle, that is cosθ Ê 0, required angle between given surfaces is θ = cos−1(

8

3p

21

).

Illustration 9.10 Find the values of constantsλ and µ so that the surfacesλx2−µy z = (λ+2) x and 4x2 y+z3 = 4 may intersect orthogonally at the point (1,−1,2) .

Solution: Given surfaces are intersect orthogonally ii angle between them at the point (1,−1,2) is θ = 90.Hence,

cosθ = 0 ⇔−→N 1 ·−→N 2∣∣∣−→N 1

∣∣∣ ∣∣∣−→N 2

∣∣∣ = 0 ⇔ −→N 1 ·−→N 2 = 0 (9.11)

Let φ1 =λx2 −µy z − (λ+2) x, φ2 = 4x2 y + z3 −4, p (1,−1,2)

∇φ1 = ∂φ1

∂xi + ∂φ1

∂yj + ∂φ1

∂zk = (2λx −λ−2) i −µz j −µyk

∇φ2 = ∂φ2

∂xi + ∂φ2

∂yj + ∂φ2

∂zk = 8x yi +4x2 j +3z2k

At the point p (1,−1,2) , normal to the given surfaces are

−→N 1 =

(∇φ1)

p = (λ−2) i −2µ j +µk and−→N 2 =

(∇φ2)

p =−8i +4 j +12k

From (9.11), −→N 1 ·−→N 2 = 0 ⇔ 8λ−4µ= 16 (9.12)

Also point p (1,−1,2) lies on both surfaces, and hence it satisfies both equations of surfaces. Putting thepoint in the surface equation λx2 −µy z = (λ+2) x, we get µ= 1. Substituting µ= 1 in (9.12), we get λ= 5/2.

Hence, required values are λ= 5

2µ= 1





Exercise 9.1

1. If r = ∣∣−→r ∣∣ , where −→r = xi + y j + zk, prove that

a. ∇ logr =−→rr 2 b. ∇

(er 2

)= 2er 2−→r

2. Find the unit normal vector to the surface,

a. x2 y +2xz = 4 at (2,−2,3) [Winter-2015] b. z = 4(x2 + y2) at (1,0,2) [Summer-2016]

3. If−→F = (

y2 − z2 +3y z −2x)

i + (3xz +2x y

)j + (

3x y −2xz +2z)

k, then show that−→F is both solenoidal

and irrotational. [summer-2016]

4. If−→F = (

x + y +1)

i + j − (x + y

)k, show that

−→F ·curl

−→F = 0.

5. If φ= x y z −2y2z +x2z2, find div(gradφ

)at the point (2,4,1) . [summer-2015]

[Hint: div(gradφ

)=∇· (∇φ)= (∇·∇)φ=∇2φ=(∂2

∂x2 + ∂2

∂y2 + ∂2

∂z2

)φ=

(∂2φ

∂x2 + ∂2φ

∂y2 + ∂2φ

∂z2

)]

6. Show that the vector field−→V = (

sin y + z)

i + (x cos y − z

)j + (

x − y)

k is irrotational. Hence find itsscalar potential function.

7. Find the constants a,b,c so that the vector−→A = (

x +2y +az)

i + (bx −3y − z

)j + (

4x + c y +2z)

k is

irrotational. If−→A = gradφ, show that φ= x2

2− 3y2

2+ z2 +2x y +4xz − y z.

8. Find the directional derivative of the following functions:

a. f = 2x y + z2 at the point (1, - 1,3) in the direction of the vector i +2 j +2k.

b. φ= 4xz3 −3x2 y z2 at the point (2,−1,2) along the Z -axis.

9. Find the directional derivative of the function f = x2 − y2 +2z2 at the point P (1,2,3) in the directionof the line PQ where Q is the point (5,0,4) .

[Hint:−−→PQ =Q −P = (5,0,4)− (1,2,3) = (4,−2,1) = 4i −2 j + k]

10. In what direction from (3.1.−2) is the directional derivative of φ= x2 y2z4 is maximum and what is itsmagnitude ?

11. What is the greatest rate of the increase of v = x2 + y z2 at the point(1,−1,3)?

12. If θ is the acute angle between the surfaces x y2z = 3x + z2 and 3x2 − y2 +2z = 1 at the point (1,−2,1),

show that cosθ = 3

7p

6.

13. Calculate the angle between the normals to the surface x y = z2 at the points (4,1,2) and (3,3,−3) .

14. Find the angle between the tangent planes to the surfaces x log z = y2 −1 and x2 y = 2− z at the point(1,1,1) .

Answers

2. a.1

3

(−i +2 j +2k)

b.1p65

(−8i + k)

5. 14 6. x sin y +xz − y z 8. a.14

3b. 144

9.28p21

10.1p19

(−i +3 j −3k)

,96p

19 11. 11 13. cos−1(

1p22

)14. cos−1

(1p30

)E E E





Powered by





Contact: (+91) 9427 80 9779





Chapter 10Vector Calculus II: Vector Integration

10.1 Line Integral

â Any integral that is evaluated along the curve is called line integral.

â Let−→F = F1 i +F2 j +F3k be the vector point function defined along the smooth curve C then integral

of−→F along C is called line integral and is defined as∫

C

−→F ·d−→r =

∫C

(F1 i +F2 j +F3k

) · (d xi +d y j +d zk)

∴∫

C

−→F ·d−→r =

∫C

(F1d x +F2d y +F3d z

)* Important:

1. If C is the closed curve then the line integral is denoted by the symbol∮

C

−→F ·d−→r .

2. Work: If−→F denotes the force acting on a moving particle along the curve C then work done by the

force is given by W =∫

C

−→F ·d−→r .

3. Circulation: If−→V denotes the velocity of the moving particle around the closed curve C then the

circulation of the particle round the curve C is defined as ω=∮

C

−→V ·d−→r .

Further if the circulation is zero then the motion is called irrotational.

4. Path independence of line Integral: If−→F is an irrotatinal vector field, that is curl

−→F =∇×−→

F =−→0 , then

the value of line integral∫

C

−→F ·d−→r does not depends on path (line integral is independent to path).

That is it depends on initial and final points.

Also, since−→F is an irrotational there exist a scalar φ such that

−→F = gradφ=∇φ

= ∂φ

∂xi + ∂φ

∂yj + ∂φ

∂zk

∴−→F ·d−→r =

(∂φ

∂xi + ∂φ

∂yj + ∂φ

∂zk

)· (d xi +d y j +d zk

) [∵−→r = xi + y j + zk

]= ∂φ

∂xd x + ∂φ

∂yd y + ∂φ

∂zd z

= dφ[∵ Total differentiation of φ

]

114




Hence line integral of−→F along any path C joining the points A and B is

∫C

−→F ·d−→r =

∫ B

Adφ= [

φ]B

A = (φ

)B − (

φ)

A

5. Arc length: The arc length of the curve defined by position vector −→r (t ) = f (t ) i + g (t ) j +h (t ) k fromt = a to t = b is given by

L =∫ b

a

∣∣∣−→r ′(t )

∣∣∣d t =∫ b

a

√[f ′ (t )

]2 + [g ′ (t )

]2 + [h′ (t )]2d t

Illustration 10.1 Evaluate the line integral∮

C

−→F ·d−→r , where

−→F = (

x2 +x y)

i + (x2 + y2) j and C is the

square formed by the lines x =±1, y =±1.

Solution: Since,∮C

−→F ·d−→r =

∮C

[(x2 +x y

)d x + (

x2 + y2) d y]

(10.1)

Here given closed curve C is the square formed by x =±1, y =±1.It is bounded by four different sub-curves curves like C1,C2,C3

C4 as shown in figure.

In order to find line integral (10.1), we will find line integralalong each sub-curve, in anti-clockwise direction (counterclockwise direction) and then will add them.

Along C1: x = 1, ∴ d x = 0 and y varies from −1 to 1.From (10.1),∫

C1

~F ·d~r =∫ 1

−1

[(1+ y

)(0)+ (

1+ y2)d y]= ∫ 1

−1

[(1+ y2)d y

]=

[y + y3

3

]1

−1=

[1+ 1

3

]−

[−1− 1

3

]= 2+ 2

3

∴∫

C1

~F ·d~r = 8

3

Along C2: y = 1, ∴ d y = 0 and x varies from 1 to −1.From (10.1),∫

C2

~F ·d~r =∫ −1

1

[(x2 +x

)d x + (

x2 +1)

(0)]= ∫ −1

1

[(x2 +x

)d x

]=

[x3

3+ x2

2

]−1

1=

[−1

3+ 1

2

]−

[1

3+ 1

2

]=−2

3

∴∫

C2

~F ·d~r =−2

3

Along C3: x =−1, ∴ d x = 0 and y varies from 1 to −1.From (10.1),∫

C3

~F ·d~r =∫ −1

1

[(1− y

)(0)+ (

1+ y2)d y]= ∫ −1

1

[(1+ y2)d y

]=

[y + y3

3

]−1

1=

[−1− 1

3

]−

[1+ 1

3

]=−2− 2

3





∴∫

C3

~F ·d~r =−8

3

Along C4: y =−1, ∴ d y = 0 and x varies from −1 to 1.From (10.1)∫

C4

~F ·d~r =∫ 1

−1

[(x2 −x

)d x + (

x2 +1)

(0)]= ∫ 1

−1

[(x2 −x

)d x

]=

[x3

3− x2

2

]1

−1=

[1

3− 1

2

]−

[−1

3− 1

2

]= 2

3

∴∫

C4

~F ·d~r = 2

3

Hence required line integral,∮C

−→F ·d−→r =

∫C1

−→F ·d−→r +

∫C2

−→F ·d−→r +

∫C3

−→F ·d−→r +

∫C4

−→F ·d−→r

= 8

3− 2

3− 8

3+ 2

3

∴∮

C

−→F ·d−→r = 0

Illustration 10.2 Find the work done when the force−→F = (

x2 − y2 +x)

i −(2x y + y

)j moves the particle in

the xy-plane from (0,0)to (1,1) along the parabola y2 = x. Is the work done different when the parh is thestraight liney = x ? [Winter-2016]

Solution: Work done by the force−→F along the path C is given by

W =∫

C

−→F ·d−→r =

∫C

[(x2 − y2 +x

)d x − (

2x y + y)

d y]

(10.2)

Along the parabola C1: y2 = x, ∴ d x = 2d y and y varies from 0to 1.From (10.2),

W =∫ 1

0

[(y4 − y2 + y2)(2yd y

)− (2y3 + y

)d y

]=

∫ 1

0

(2y5 −2y3 − y

)d y

=[

2y6

6−2

y4

4− y2

2

]1

0= 1

3− 1

2− 1

2∴W =−2

3

Work done along the parabola y2 = x is − 2

3.

Along the line C2: y = x, ∴ d x = d y and y varies from 0 to 1.From (10.2),

W =∫ 1

0

[(y2 − y2 + y

)(d y

)− (2y2 + y

)d y

]=

∫ 1

0

(−2y2)d y =[−2

y3

3

]1

0=−2

3∴W =−2

3

Work done along the line y = x is also − 2

3.

Hence work is not different for given different paths.Note that here line integration is independent of path. Also such a system in which work does not

depends on path is called conservative system.





Illustration 10.3 If−→F = (

2x y + z3) i +x2 j +3xz2k, show that∫

C

−→F ·d−→r is independent of path of integra-

tion. Hence find the integral when C is any path joining (1,−2,1) and (3,1,4) . [Summer-2016]

Solution: We know that line integral of−→F is independent of path, if the vector field is irrotational. That is

curl~F =−→0 .

Given−→F = (

2x y + z3) i +x2 j +3xz2k,

curl~F =∇×~F =

∣∣∣∣∣∣∣∣∣∣∣i j k∂

∂x

∂

∂y

∂

∂z(2x y + z3) x2 3xz2

∣∣∣∣∣∣∣∣∣∣∣= i

[∂

∂y

(3xz2)− ∂

∂z

(x2)]− j

[∂

∂x

(3xz2)− ∂

∂z

(2x y + z3)]+ k

[∂

∂x

(x2)− ∂

∂y

(2x y + z3)]

= i [0−0]− j[3z2 −3z2]+ k [2x −2x] =−→

0 ∴ curl~F =−→0

Now in order to find line integral, let

−→F = gradφ= ∂φ

∂xi + ∂φ

∂yj + ∂φ

∂zk ⇒ ∂φ

∂x= 2x y + z3,

∂φ

∂y= x2,

∂φ

∂z= 3xz2

Integrating above equations partially with respect to x, y, z respectively, keeping other variable constant, weget following equations: [See Illustration 9.7]

φ= x2 y +xz3 + c1(y, z

), φ= x2 y + c2 (x, z) , φ= xz3 + c3

(x, y

) ⇒ φ= x2 y +xz3

Hence required path independent line integral joining the points A (1,−2,1) and B (3,1,4) is given by∫C

−→F ·d−→r =

∫ B

Adφ= [

φ]B

A = [φ

]B − [

φ]

A

= [x2 y +xz3]

(3,1,4) −[x2 y +xz3]

(1,−2,1)

= [201]− [−1] = 202

∴∫

C

−→F ·d−→r = 202 Ans.

Illustration 10.4 Find the arc length of the portion of the circular helix −→r (t ) = cos t i + sin t j + t k fromt = 0 to t =π. [Summer-2015]

Solution: Given~r (t ) = cos t i + sin t j + t k ⇒ ~r ′ (t ) = d~r

d t=−sin t i +cos t j + k, 0 É t Éπ

By definition of arc length,

L =∫ π

0

∣∣~r ′∣∣d t =∫ π

0

√(−sin t )2 + (cos t )2 +1d t

=∫ π

0

√sin2t +cos2t +1d t =

∫ π

0

p2d t

=[p

2t]π

0=p

2π ∴ L =p2π Ans.





10.2 Surface Integral (Normal Surface Integral)

â Any integral that is evaluated over the surface is called line inte-gral.

â Let−→F = F1 i + F2 j + F3k be the vector point function defined

over the smooth surface S and R be the orthogonal projectionon x y− plane. If n denotes the unit outward normal vector to

the surface S then the surface integral of−→F over S is defined as

ÏS

−→F · n d s =

ÏR

−→F · n

d xd y∣∣ n · k∣∣

Remark:

1. Surface integral is denoted by the symbolÏ

S

−→F ·−→d s OR

∫S

−→F ·−→d s OR

∫S

−→F · n d s.

2. Instead of on x y−plane, if we take orthogonal projection of surface S on y z−plane or zx−plane, then

surface integrals areÏ

S

−→F · n d s =

ÏR

−→F · n

d yd z∣∣ n · i∣∣ or

ÏS

−→F · n d s =

ÏR

−→F · n

d xd z∣∣ n · j∣∣ respectively.

3. Flux: If−→F denotes the velocity of the fluid then

ÏS

−→F · n d s gives the amount of fluid emerging from

the surface area S per unit time, that is it gives flux.

Further ifÏ

S

−→F · n d s = 0, then

−→F is called solenoidal.

Illustration 10.5 EvaluateÏ

S

−→F · nd s, where

−→F = 18zi −12 j +3yk and S is the surface of the plane 2x +

3y +6z = 12 in the positive octant.

Solution:Given surface S : 2x + 3y + 6z = 12, that is

x

6+ y

4+ z

2= 1, is a

plane in first octant with intercepts (6,0,0) , (0,4,4) and (0,0,2)on x, y and z axis respectively. Let R be orthogonal projection onx y−plane as show in figure. By definition of surface integralÏ

S

−→F · nd s =

ÏR

−→F · n

d xd y∣∣n · k∣∣ (10.3)

where n is unit outward normal vector to the given surface S :2x +3y +6z = 12, and is given by [See section 9.5]

n = ∇S

|∇S| =2i +3 j +6kp

22 +32 +62= 1p

7

(2i +3 j +6k

)∴

∣∣n · k∣∣= ∣∣∣∣6

7

∣∣∣∣= 6

7and

−→F · n = 1

7

(2i +3 j +6k

) · (18zi −12 j +3yk)= 1

7

(36z −36+18y

)Since R is projection on x y−plane, that is RHS of (10.5) is a double integral in x, y , we convert

−→F · n in terms

of x, y bu putting the value of z from equation of surface 2x +3y +6z = 12. That is put z = 1

6

(12−2x −3y

).

∴−→F · n = 1

7

[6(12−2x −3y

)−36+18y]= 6

7(6−2x)

Substituting values in (10.5), we getÏS

−→F · nd s =

ÏR

6

7(6−2x)

d xd y(67

) =Ï

R(6−2x)d xd y





where projection R is triangle in x y−plane as show in figure. For limits of double integral, according to

Y −strip (parallel to Y −axis), 0 É y É 1

3(12−2x) , 0 É x É 6. Hence required surface integral,

ÏS

−→F · nd s =

∫ 6

0

∫ 13 (12−2x)

0(6−2x)d yd x

=∫ 6

0(6−2x)

[y] 1

3 (12−2x)0 d x

[∵ Integrating w.r.t. y keeping x constant

]=

∫ 6

0(6−2x)

[1

3(12−2x)−0

]d x

= 4

3

∫ 6

0(3−x) (6−x)d x = 4

3

∫ 6

0

(18−9x +x2)d x

= 4

3

[18x − 9x2

2+ x3

3

]6

0= 4

3[18−0]

∴Ï

S

−→F · nd s = 24 Ans.

Illustration 10.6 Show thatÏ

S

−→F · n d s = 3

2, whare

−→F = 4xzi − y2 j + y zk and S is the surface of the cube

bounded by the planes x = 0, x = 1, y = 0, y = 1, z = 0, z = 1.

Solution:Given S is the closed surface of unit cube. It isbounded by six different subsurface (squares) asshow in figure, like

S1 : x = 0 (OC DE square in the y z−plane)

S2 : x = 1 (ABGF square parallel to y z−plane)

S3 : y = 0 (OEF A square in the xz−plane)

S4 : y = 1 (C DGB square parallel xz−plane)

S5 : z = 0 (O ABC square in the x y−plane)

S5 : z = 1 (EFGD square parallel to x y−plane)

In order to find surface integral of−→F = 4xzi − y2 j + y zk we will find surface integral over each subsurface

and then will add them.Over S1: x = 0, and since S1 lies in y z−plane, the unit outward normal vector (in outside direction of cube,that is unit vector along negative X−axis direction) is n =−i .

∴−→F · n = (

4xzi − y2 j + y zk) · (−i

)=−4xz = 0 [∵ x = 0] ⇒Ï

S1

−→F · nd s =

ÏS1

(0)d s = 0

Over S2: x = 1, and since S1 is parallel to y z−plane, the unit outward normal vector (in outside directionof cube, that is unit vector along positive X−axis direction) is n = i .

∴−→F · n = (

4xzi − y2 j + y zk) · (i

)= 4xz = 4z [∵ x = 1] ⇒Ï

S2

−→F · nd s =

ÏS2

4zd s

Since S2 is parallel to y z−plane, its orthogonal projection is on the y z−plane and it is a square OC DE .(0 É y É 1, 0 É z É 1

)We put formula of surface integral for y z−plane.

∴Ï

S2

−→F · nd s =

ÏS2

4zd s = 4∫ 1

0

∫ 1

0z

d yd z∣∣n · i∣∣





= 4∫ 1

0

∫ 1

0zd yd z

[∵ n · i = 1

]= 4

[y]1

0

[z2

2

]1

0= 4[1]

[1

2

]= 2 ∴

ÏS2

−→F · nd s = 2

Similarly, we have other integrals as follow:

Over S3: y = 0, n =− j ⇒ −→F · n = y2 = 0

∴Ï

S3

−→F · nd s =

ÏS3

(0)d s = 0

Over S4: y = 1, n = j ⇒ −→F · n =−y2 =−1

∴Ï

S4

−→F · nd s =

ÏS4

(−1)d s =−∫ 1

0

∫ 1

0d xd z =−1 ∴

ÏS4

−→F · nd s =−1

Over S5: z = 0, n =−k ⇒ −→F · n =−y z = 0

∴Ï

S5

−→F · nd s =

ÏS5

(0)d s = 0

Over S6: z = 1, n = k ⇒ −→F · n = y z = y

∴Ï

S6

−→F · nd s =

ÏS4

(y)

d s =∫ 1

0

∫ 1

0yd xd y = 1

2∴

ÏS6

−→F · nd s = 1

2

Hence, ÏS

−→F · nd s =

ÏS1

−→F · nd s +

ÏS2

−→F · nd s +

ÏS3

−→F · nd s +

ÏS4

−→F · nd s +

ÏS5

−→F · nd s +

ÏS6

−→F · nd s

= 0+2+0−1+0+ 1

2

∴Ï

S

−→F · nd s = 3

2Proved.

10.3 Volume Integral

â Any integral that is evaluated through the volume of solid is called volume integral.

â Let−→F = F1 i +F2 j +F3k be the vector point function andφ

(x, y, z

)be the scalar point function defined

through the volume V of the solid then the volume integrals are defined asÑV

−→F d v OR

ÑVφd v , where d v = d xd yd z

Exercise 10.1

1. If−→F = 3x yi − y2 j . evaluate

∫C

−→F ·d−→r where C is the curve y = 2x2 from (0,0) to (1,2) .

2. Evaluate∮

C

−→F .

−→dr , where

−→F = ex sin yi +ex cos y j and C is the rectangle whose vertices are (0,0) , (1,0) ,(

1,π

2

), and

(0,π

2

).

3. Find the circulation of−→F round the curve C, that is

∮C

−→F .

−→dr where

−→F = yi +z j +xk and C is the circle

x2 + y2 = 1 and z = 0.

[Hint: Let x = cosθ, y = sinθ, z = 0 0 É θ É 2π⇒ d x =−sinθdθ, y = cosθdθ,d z = 0]





4. Find the work done in moving a particle by the force field−→F = 3x2 i+(

2xz − y)

j+zk, along the straightline from (0,0,0) to (2,1,3) .

5. Determine the length of curve −→r (t ) = 2t i +3sin(2t ) j +33cos(2t ) k. on the interval 0 É t É 2π.

6. If−→F = 2x y z3 i + x2z3 j +3x2 y z2k, show that

∫C

−→F ·d−→r is independent of path of integration. Hence

find the integral when C is any path joining (0,0,0) and (1,2,3) .

Further show that for any simple closed curve C ,∮

C

−→F ·d−→r = 0.

7. EvaluateÏ

S

−→F · nd s, where

−→F = xi+xz j+4x yk and S is the triangular surface with the vertex (2,0,0) , (0,2,0)

and (0,0,4) .

[Hint: S :x

2+ y

2+ z

4= 1 i.e. S : 2x +2y + z = 4 See Illustration 10.5]

8. EvaluateÏ

S

−→F · n d s, whare

−→F = y zi +zx j +x yk and S is that part of the surface x2+y2+z2 = 1 which

lies in the positive octant.

9. Show that∮

C

−→r ·d−→r = 0 independently of the origin of −→r .

10. Evaluate∫

C

(yd x +xd y + zd z

), where C is given by x = cos t , y = sin t , z = t 2,0 É t É 2π. [Winter-2015]

Answers

1. −7

62. 0 3. −π 4. 16 5. 4π

p10 6. 54 7. 8 8.

3

89.

E E E

10.4 Integral Theorems

Theorem 10.1 (Green’s Theorem1 : Relation between Line Integral & Double Integral)Let M

(x, y

)and N

(x, y

)are functions of two variable having

continuous first order partial derivative in the region R ofx y−plane bounded by the closed curve C then

∮C

(Md x +N d y

)=ÏR

(∂N

∂x− ∂M

∂y

)d xd y

where C traverse in anti-clockwise direction.

Illustration 10.7 Verify Green’s theorem in the plane for∮

C

(3x −8y2)d x + (

4y −6x y)

d y , where C is the

boundary of the boundary of the triangle with vertices (0,0) , (1,0) and (0,1) . [Summer-2017]

Solution: By Green’s theorem, ∮C

(Md x +N d y

)=ÏR

(∂N

∂x− ∂M

∂y

)d xd y

Here M = (3x −8y2), N = (

4y −6x y)

and C is the triangle with vertices (0,0) , (1,0) , (0,1) . That is trianglebounded by x = 0, y = 0, x + y = 1 as show in figure.

1George Green; English, 1793-1841.





To find line integral:∮C

Md x +N d y =∮

C

(3x −8y2)d x + (

4y −6x y)

d y (10.4)

Along C1: y = 0 ⇒ d y = 0 x : 0 → 1

From (10.4), I1 =∫ 1

0[(3x −0)d x +0] =

[3x2

2

]1

0= 3

2

Along C2: x + y = 1 ∴ y = 1−x ⇒ d y =−d x; x : 1 → 0

From (10.4), I2 =∫ 0

1

[3x −8(1−x)2d x + 4(1−x)−6x (1−x) (−d x)

]=−

∫ 1

0

[3x −8

(1−2x +x2)− 4(1−x)−6x (1−x)

]d x

=−∫ 1

0

[3x −8+16x −8x2 −4+4x +6x −6x2]d x

=−∫ 1

0

[−14x2 +29x −12]

d x =−[−14x3

3+ 29x2

2−12x

]1

0=−

[−14

3+ 29

2−12

]∴ I2 = 13

6Along C3: x = 0 ⇒ d x = 0; y : 1 → 0

From (10.4), I3 =∫ 0

1

[0+ (

4y −0)

d y]= ∫ 0

14yd y = [

2y2]01 =−2

Hence,∮C

Md x +N d y =I1 + I2 + I3 = 3

2+ 13

6−2 = 3

3(10.5)

To find double integral: Here M = 3x −8y2, N = 4y −6x y ⇒ ∂M

∂y=−16y,

∂N

∂x=−6y

∴∂N

∂x− ∂M

∂y=−6y +16y = 10y

ÏR

(∂N

∂x− ∂M

∂y

)d xd y =

ÏR

(10y

)d xd y = 10

ÏR

yd xd y

where R is triangular region as show in figure. According to Y −strip (parallel to Y − axis), limits of doubleintegral are 0 É y É 1−x, 0 É x É 1.Ï

R

(∂N

∂x− ∂M

∂y

)d xd y = 10

∫ 1

0

∫ 1−x

0ydd x y

= 10∫ 1

0

[y2

2

]1−x

0d x = 10

∫ 1

0

[(1−x)2

2

]d x

= 5∫ 1

0(1−x)2d x = 5

[(1−x)3

3(−1)

]1

0=−5

3[0−1]

∴Ï

R

(∂N

∂x− ∂M

∂y

)d xd y = 5

3(10.6)

Hence from (10.5) and (10.6), Green’s theorem is verified.

Illustration 10.8 Apply the Green’s theorem to evaluate∮

C

(y − sin x

)d x +cos x d y

where C is the plane

triangle enclosed by the lines y = 0, x = π2 and y = 2

πx.

Solution: By Green’s theorem,∮C

(Md x +N d y

)=ÏR

(∂N

∂x− ∂M

∂y

)d xd y (10.7)





Here M = (y − sin x

), N = cos x and C is boundary of triangular

region R enclosed by the lines y = 0, x = π2 and y = 2

πx as show infigure.

To evaluate line integral using Green’s theorem, we find doubleintegral. According to Y −strip limits of double integral are

0 É y É 2x

π, 0 É x É π

2.

From (10.7), required line integral

∮C

(Md x +N d y

)=ÏR

(−sin x −1)d xd y

=−∫ π/2

0

∫ 2x/π

0(sin x +1)d yd x

=−∫ π/2

0(sin x +1)

[y]2x/π

0 d x =−∫ π/2

0(sin x +1)

[2x

π−0

]d x

=− 2

π

∫ π/2

0x (sin x +1)d x =− 2

π

∫ π/2

0(x sin x +x)d x

=− 2

π

[(x) (cos x)− (1)(−sin x)+ x2

2

]π/2

0

[∵ Integrating by parts

]=− 2

π

[(π2

)(cos

π

2

)+

(sin

π

2

)+ 1

2

(π2

)2−0

]=− 2

π

[0+1+ π

8

2]

∴∮

C

(y − sin x

)d x +cos xd y

=− 2

π

(0+1+ π

8

)Ans.

Illustration 10.9 Apply Green’s theorem to prove area encased by plane curve is1

2

∮C

(xd y − yd x

). Hence,

find the area of an ellipse whose semi-major and semi-minor axes are of length a and b.

Solution: Let R be the region enclosed by simple closed curve C , then by Green’s theorem∮C

(Md x +N d y

)=ÏR

(∂N

∂x− ∂M

∂y

)d xd y

From given line integral, M =−y, N = x ⇒ ∂N

∂x− ∂M

∂y= 1− (−1) = 2

∮C

(xd y − yd x

)= 2Ï

Rd xd y = 2

(Area enclosed by closed curve C

)∴ Area enclosed by closed curve C = 1

2

∮C

(xd y − yd x

)

Now equation of given ellipse is C :x2

a2 +y2

b2 = 1. To find line integral substitute parametric equation of ellipse

in above formula, that is

x = a cos t , y = b sin t , 0 É t É 2π ⇒ d x =−a sin t , d y = b cos t

A = 1

2

∮C

(xd y − yd x

)= 1

2

∫ 2π

0[(a cos t ) (b cos td t )− (b sin t ) (−a sin td t )]





= 1

2

∫ 2π

0

(abcos2t +absin2t

)d t = ab

2

∫ 2π

0

(cos2t + sin2t

)d t

= ab

2

∫ 2π

0(1)d t = ab

2[t ]2π

0 = ab

2[2π]

∴ A =πab Ans.

Note: To find an area of the circle of radius a, take a = b in above illustration. (Verify !)

Theorem 10.2 (Stokes’ theorem2 : Relation between Line Integral & Surface Integral)

Let−→F be the differentiable vector point function defined over an

open surface S bounded by the closed curve C. If n denotes theunit outward normal vector to the surface S, then∮

C

−→F ·d−→r =

ÏS

culr−→F · nd s =

ÏS

(∇×−→

F)· nd s

where C traverse in anti-clockwise direction.

Illustration 10.10 Verify Stokes’ theorem for the vector field−→F = (

y − z +2)

i + (y z +4

)j − xzk over the

box bounded by the planes x = 0, y = 0, z = 0, x = 2, y = 2, z = 2 above the xy-plane. [Summer-2015]

Solution: Stokes’ theorem:∮

C

~F ·d~r =Ï

Sculr~F · nd s

where S is the surface of the box bounded by the planes x = 0, y = 0, z = 0, x = 2, y = 2, z = 2 above the xy-plane (that is open at the bottom) and C is the boundary of the square in x y−plane (z = 0 plane) as shownin figure.To find surfave integral:

Consider five sub-surfaces,

S1 : x = 0 (OC DE square in the y z−plane)

S2 : x = 2 (ABGF square parallel to y z−plane)

S3 : y = 0 (OEF A square in the xz−plane)

S4 : y = 2 (C DGB square parallel xz−plane)

S5 : z = 2 (EFGD square parallel to x y−plane)

Here−→F = (

y − z +2)

i + (y z +4

)j −xzk

∴ curl−→F =∇×−→

F =

∣∣∣∣∣∣∣∣∣∣∣i j k∂

∂x

∂

∂y

∂

∂z

y − z +2 y z +4 −xz

∣∣∣∣∣∣∣∣∣∣∣= i

[∂

∂y(−xz)− ∂

∂z

(y z +4

)]− j

[∂

∂x(−xz)− ∂

∂z

(y − z +2

)]+ k

[∂

∂x

(y z +4

)− ∂

∂y

(y − z +2

)]= i

[0− y

]− j [−z − (−1)]+ k [0− (1)]

∴ curl−→F =−yi + (z −1) j − k

2Sir George Stokes; Irish, 1819-1903.





Over S1: x = 0, n =−i ⇒ curl−→F · n = y

∴Ï

S1

curl−→F · nd s =

∫ 2

0

∫ 2

0yd xd y =4

Over S2: x = 2, n = i ⇒ curl−→F · n =−y

∴Ï

S2

curl−→F · nd s =−

∫ 2

0

∫ 2

0yd xd y =−4

Over S3: y = 0, n =− j ⇒ curl−→F · n = (z −1)

∴Ï

S3

curl−→F · nd s =−

∫ 2

0

∫ 2

0(z −1)d xd z =0

Over S4: y = 2, n = j ⇒ curl−→F · n = (z −1)

∴Ï

S3


∫ 2

0

∫ 2

0(z −1)d xd z =0

Over S5: z = 2, n = k ⇒ curl−→F · n =−1

∴Ï

S5


∫ 2

0

∫ 2

0(−1)d xd y =−4

Hence, ÏS


Ï(S1+S2+S3+S4+S5)

curl−→F · nd s = 4−4+0+0−4

∴Ï

Scurl

−→F · nd s =−4 (10.8)


−→F ·−→dr =

∮C

[(y − z +2

)d x + (

y z +4)

d y −xzd z]= ∮

C

[(y +2

)d x +4d y

][∵ z = 0] (10.9)

Along C1: y = 0 ⇒ d y = 0, x : 0 → 2

∴∫

C1

−→F ·−→dr =

∫ 2

02d x = 4

Along C2: x = 2 ⇒ d x = 0, y : 0 → 2

∴∫

C2

−→F ·−→dr =

∫ 2

04d y = 8

Along C3: y = 2 ⇒ d y = 0, x : 2 → 0

∴∫

C3

−→F ·−→dr =

∫ 0

24d x =−8

Along C4: x = 0 ⇒ d x = 0, y : 2 → 0

∴∫

C4

−→F ·−→dr =

∫ 0

24d y =−8

Hence from (10.9),∮C

−→F ·−→dr =

∫C1

−→F ·−→dr +

∫C2

−→F ·−→dr +

∫C3

−→F ·−→dr +

∫C3

−→F ·−→dr

= 4+8−8−8 =−4

∴∮

C

−→F ·−→dr =−4 (10.10)

Thus from (10.8) and (10.10), Stokes’ theorem is verified.





Illustration 10.11 Verify Stokes’ theorem for the vector field−→F = (

2x − y)

i − y z2 j − y2zk over the upperhalf surface of the sphere x2 + y2 + z2 = 1 and C is its boundary.

Solution: Stokes’ theorem:∮

C

~F ·d~r =Ï

Sculr~F · nd s

where S is the surface of unit sphere x2 + y2 + z2 = 1 above the xy-plane (that is open at the bottom) and Cis the boundary of the unit circle x2 + y2 = 1 in x y−plane (z = 0 plane) as shown in figure.To find surfce integral:

Here−→F = (

2x − y)

i − y z2 j − y2zk

∴ curl−→F =∇×−→

F =

∣∣∣∣∣∣∣∣∣∣∣i j k∂

∂x

∂

∂y

∂

∂z

2x − y −y z2 −y2z

∣∣∣∣∣∣∣∣∣∣∣= (−2y z +2y z

)i − (0−0) j + (0+1) k = k

∴ curl−→F = k

Also n = ∇S

|∇S| =(2xi +2y j +2zk

)√4x2 +4y2 +4z2

[∵ S : x2 + y2 + z2 = 1

]=

(2xi +2y j +2zk

)2√

x2 + y2 + z2=

(2xi +2y j +2zk

)2p

1

∴ n = xi + y j + zk

⇒ curl−→F · n = k · (xi + y j + zk

)= z,∣∣n · k

∣∣= |z| = zÏS


ÏR

curl−→F · n

d xd y∣∣n · k∣∣

where R is region of bounded by circle x2 + y2 = 1 in x y −plane.

=Ï

R6 z d xd y

6 z =Ï

Rd xd y

= Area enclosed by R : x2 + y2 = 1 [∵ Formula of area]

=π(radius)2 =π(1)2 =π∴

ÏS

curl−→F · nd s =π (10.11)


−→F ·−→dr =

∮C

[(2x − y

)d x − y z2d y − y2zd z

]=

∮C

(2x − y

)d x [∵ z = 0]

Since C is the boundary of the circle x2 + y2 = 1, using parametric substitution

x = cos t , y = sin t , 0 É t É 2π ⇒ d x = sin t , d y = cos t

We get, ∮C

−→F ·−→dr =

∫ 2π

0(2cos t − sin t ) (−sin td t )

=∫ 2π

0

(−2sin t cos t + sin2t)

d t =∫ 2π

0

(−sin2t + 1−cos2t

2

)d t





=[

cos2t

2+ 1

2t − sin2t

4

]2π

0

=[

cos4π

2+ 1

2(2π)− sin4π

4

]−

[1

2+0−0

]=

[1

2+π−0

]−

[1

2

]=π

∴∮

C

−→F ·−→dr =π (10.12)

Thus from (10.11) and (10.12), Stokes’ theorem is verified.

Theorem 10.3 (Gauss Divergence Theorem3 : Relation between Surface Integral & Volume Integral)

Let−→F be the differentiable vector point function defined

through the volume V enclosed by the closed surface S. If n de-notes the unit outward normal vector to the surface S, thenÏ

S

−→F · nd s =

ÑV

div−→F d v =

ÑV∇·−→F d v

Illustration 10.12 Verify the divergence theorem for−→F = 4xzi − y2 j + y zk taken over the cube bounded

by the planes x = 0, x = 1, y = 0, y = 1, z = 0, z = 1.

Solution: Divergence theorem:Ï

S

−→F · nd s =

ÑV

div−→F d v =

ÑV

(∇·−→F

)d xd yd z

To find surface integral:

Here−→F = 4xzi − y2 j + y zk and S is the closed surface of unit cube. [See Illustration 10.6]

∴Ï

S

−→F · nd s = 3

2(10.13)

To find volume integral

−→F = 4xzi − y2 j + y zk

∴ div−→F =∇·−→F = ∂

∂x(4xz)+ ∂

∂y

(−y2)+ ∂

∂z

(y z

)= 4z −2y + y = 4z − y

Also V is the volume of unit cube. So limits of triple integral are 0 É x É 1, 0 É y É 1, 0 É z É 1.ÑV

div−→F d v =

∫ 1

0

∫ 1

0

∫ 1

0

(4z − y

)d xd yd z

=∫ 1

0

∫ 1

0

(4z − y

)[x]1

0 d yd z =∫ 1

0

∫ 1

0

(4z − y

)d yd z

=∫ 1

0

[4y z − y2

2

]1

0d z =

∫ 1

0

(4z − 1

2

)d z

=[

2z2 − 1

2z

]1

0= 2− 1

2

∴Ñ

Vdiv

−→F d v = 3

2(10.14)

Hence from (10.13) and (10.14), Gauss divergence theorem is verified.

3Johann Carl Friedrich Gauss; German, 1777-1855.





Illustration 10.13 Use divergence theorem to evaluateÏ

S

−→F ·−→d s, where

−→F = x3 i + x2 y j + x2zk and S is

the surface bounding the region x2 + y2 = a2, z = 0, z = b. [Winter-2015]

Solution:Here

−→F = x3 i +x2 y j +x2zk

∴ div−→F =∇·−→F

= ∂

∂x

(x3)+ ∂

∂y

(x2 y

)+ ∂

∂z

(x2z

)= 3x2 +x2 +x2

∴ div−→F = 5x2

By Divergence theorem,ÏS

−→F nd s =

ÑV

div−→F d v

= 5Ñ

Vx2d xd y xz (10.15)

Since V is the volume of cylinder (as show in figure), for triple integral we change cartesian coordinate(x, y, z

)into cylindrical coordinate (r,θ, z) , as

x = r cosθ, y = r sinθ, z = z,

x2 + y2 = r 2, d xd yd z = r dr dθd z

0 É r É a, 0 É θ É 2π, 0 É z É b [limits for whole cylinder]

Substituting in (10.15), we getÏS

−→F nd s = 5

∫ b

0

∫ 2π

0

∫ a

0(r cosθ)2r dr dθd z

= 5

(∫ b

0d z

)(∫ 2π

0cos2θdθ

)(∫ a

0r 3dr

) [∵ Separating intgrals

]= 5[z]b

0 ×[∫ 2π

0

(1+cos2θ

2

)dθ

]×

[r 4

4

]a

0

= 5a4b

8

[θ+ sin2θ

2

]2π

0= 5a4b

8

[(2π+ sin4π

2

)−0

]∴

ÏS

−→F · nd s = 5πa4b

4Ans.

Illustration 10.14 If S is any closed surface enclosing the volume V and−→F = xi +2y j +3zk, prove thatÏ

S

−→F · nd s = 6V.

Solution: If S is a closed surface enclosing the volume V , then by Divergence theorem we haveÏS

−→F nd s =

ÑV

div−→F d v

Let−→F = xi +2y j +3zk ⇒ div

−→F =∇·−→F = 6

∴Ï

S

−→F · nd s =

ÑV

(6)d v = 6Ñ

Vd v = 6

(Volume enclosed by S

)∴

ÏS

−→F · nd s = 6V Proved.

Exercise 10.2

1. Verify Green’s theorem for the function−→F = (

x + y)

i +2x y j in the x y−plnae for the region boundedby x = 0, y = 0, x = a and y = b. [Summer-2016]





2. Verify Green’s theorem in the plane for∮

C

(3x2 −8y2)d x + (

4y −6x y)

d y , where C is the boundary of

the region bounded by y =px , y = x2.

3. Verify Green’s theorem in the plane for∮

C

(x y + y2) d x +x2 d y

, where C is the closed curve of the

region bounded y = x and y = x2.

4. Apply the Green’s theorem to evaluate∮

C

(y2d x +x2d y

)where C is the plane triangle enclosed by the

lines x = 0, y = 0 and x + y = 1. [Winter-2016]

5. Use the Green’s theorem to evaluate∮

C

−→F ·d−→r , where

−→F = −yi +x j

x2 + y2 and C is the x2 + y2 = 1 traversed

in counterclockwise direction. [Winter-2015]

6. Verify Stokes’ theorem for the vector field−→F = (

x2 + y2) i −2x y j integrated round the rectangle in thez = 0 plane and bounded by the lines x = 0, y = 0, x = a and y = b.

7. Verify Stokes’ theorem for the vector field−→F = (

x2 − y2) i +2x y j over the box bounded by the planesx = 0, x = a, y = 0, y = b, z = 0, z = c; if the face z = 0 is cut.

8. Use Stokes’ theorem to evaluateÏ

Scurl

−→F · nd s

−→F = z2 i−3x y j +x3 y3k and S is a part of z = 5−x2−y2

above the plane z = 1. Assume that S is oriented upwards.

[Hint: In this case the boundary curve C will be where the surface intersects the plane z = 1 and sowill be the curve 1 = 5−x2 − y2 ⇒ x2 + y2 = 4. ]

9. Use Stokes’ theorem to evaluate∮

C

−→F ·−→dr , where

−→F = (

x + y)

i +(2x − z) j −(y + z

)k and C is the trian-

gle with vertices (1,0,0) , (0,1,0) and (0,0,1) with counter-clockwise rotation.

10. Use divergence theorem to evaluateÏ

S


−→F = yi +x j +z2k and S is the cylindrical region

bounded by x2 + y2 = a2, z = 0 and z = h.

11. Use divergence theorem to evaluateÏ

S


−→F = x3 i + y3 j + z3k and S is the surface of the

sphere x2 + y2 + z2 = a2.

[Hint: Use spherical polar coordinate for triple integral.]

12. For any closed surface S, prove thatÏ

S

[x

(y − z

)i + y (z −x) j + z

(x − y

)k]−→

d s = 0.

[Hint: div−→F = 0]

13. If−→F = axi +by j +czk, a,b,c are constant then show that

ÏS

−→F · n d s = 4

3π (a +b + c) . Where S is the

surface of the unit sphere.

Answers

4. 0 5. 2π 8. 0 9.1

210. πa2h2 11.

12πa5

5

E E E





References:

1. Introduction to Linear Algebra with Application, Jim Defranza, Daniel Gagliardi, Tata McGraw-Hill.

2. Elementary Linear Algebra, Applications version, Anton and Rorres, Wiley India Edition.

3. Advanced Engineering Mathematics, Erwin Kreysig, Wiley Publication.

4. Higher Engineering Mathematics, B. S. Grewal, Khanna Publishers.

5. A Textbook of Engineering Mathematics, N. P. Bali, Laxmi Publications.

Powered by





Contact: (+91) 9427 80 9779



Strictly as per GTU syllabus Linear Algebra and Vector...

Documents

Transcript of Strictly as per GTU syllabus Linear Algebra and Vector...