Topic: DIAGONALIZATION,EIGEN VALUE, EIGEN VECTOR,ORTHOGONAL,CAYLEY HAMILTON THEORM

By rajesh goswami

DIAGONALIZATION EIGEN VALUE EIGEN VECTOR ORTHOGONAL CAYLEY HAMILTON THEOREM

kshumENGG2013 2

A square matrix M is called diagonalizable if we can find an invertible matrix, say P, such that the product P–1 M P is a diagonal matrix.

A diagonalizable matrix can be raised to a high power easily. › Suppose that P–1 M P = D, D diagonal. › M = P D P–1.› Mn = (P D P–1) (P D P–1) (P D P–1) … (P D P–1) = P Dn P–1.

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Let

A is diagonalizable because we can find a matrix

such that

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By definition a matrix M is diagonalizable if

P–1 M P = Dfor some invertible matrix P, and

diagonal matrix D.or equivalently,

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Suppose that

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Given a square matrix A, a non-zero vector v is called an eigenvector of A, if we an find a real number (which may be zero), such that

This number is called an eigen value of A, corresponding to the eigenvector v.

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Matrix-vector product Scalar product of a vector

If v is an eigenvector of A with eigen value , then any non-zero scalar multiple of v also satisfies the definition of eigenvector.

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k 0

First eigenvalue = 2, with eigenvector

where k is any nonzero real number.

Second eigenvalue = -5, with eigenvector

where k is any nonzero real number.

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Matrix addition/subtraction› Matrices must be of same size.

Matrix multiplication

Condition: n = q

m x n q x p m x p

Example:

2 x 2

3 x 3

n x n

diagonal matrix:

The inverse A-1 of a matrix A has the property:

AA-1=A-1A=I

A-1 exists only if

Terminology› Singular matrix: A-1 does not exist› Ill-conditioned matrix: A is close to being

singular

Properties of the inverse:

The pseudo-inverse A+ of a matrix A (could be non-square, e.g., m x n) is given by:

It can be shown that:

Equal to the dimension of the largest square sub-matrix of A that has a non-zero determinant.

Example: has rank 3

Alternative definition: the maximum number of linearly independent columns (or rows) of A.

Therefore,rank is not 4 !Example:

• A is orthogonal if:

• Notation:

Example:

• A is orthonormal if:

• Note that if A is orthonormal, it easy to find its inverse:

Property:

Diagonalize the matrix A = by means of an

orthogonal transformation.Solution:- Characteristic equation of A is

204060402

66,2,λ0λ)16(6λ)λ)(2λ)(6(2

0λ204

0λ6040λ2

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Example :-

I

1

1 2

3

1

1

2

3

1 3

2

1 3

1 1 2 3 1

1 1

xwhenλ = -2,let X = x be theeigenvector

x

then (A +2 )X = 0

4 0 4 x 00 8 0 x = 04 0 4 x 0

4x + 4x = 0 ...(1)8x = 0 ...(2)

4x + 4x = 0 ...(3)x = k ,x = 0,x = -k

1X = k 0

-1

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2

2I

0

1

2

3

1

2

3

1 3

1 3

1 3 2

2 2 3

xwhenλ = 6,let X = x be theeigenvector

x

then (A - 6 )X = 0

-4 0 4 x 00 0 x = 04 0 -4 x 0

4x + 4x = 04x - 4x = 0

x = x and x isarbitraryx must be so chosen that X and X are orthogonal among th

.1

emselvesand also each is orthogonal with X

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2 3

3 1

3 2

3

1 αLet X = 0 and let X = β

1 γ

Since X is orthogonal to Xα - γ = 0 ...(4)

X is orthogonal to Xα+ γ = 0 ...(5)

Solving (4)and(5), we get α = γ = 0 and β is arbitrary.0

Taking β =1, X = 10

1 1 0Modal matrix is M = 0 0 1

-1 1

0

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The normalised modal matrix is1 1 02 2

N = 0 0 11 1- 02 21 10 - 1 1 02 2 2 0 4 2 21 1D =N'AN = 0 0 6 0 0 0 12 2 4 0 2 1 1- 00 1 0

2 2

-2 0 0D = 0 6 0 which is the required diagonal matrix

0 0 6.

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nnnn2n1n

n22221

n11211

a...aa................a...aaa...aa

A

11 12 1n

21 22 2n

n1 n2 nn

φ(λ) = A - λIa - λ a ... a

a a - λ ... a=

... ... ... ...a a ... a - λ

| A - λI |= 0

n n-1 n-20 1 2 n

n n-1 n-20 1 2 n

We are to prove that

p λ +p λ +p λ +...+p = 0

p A +p A +p A +...+p I= 0 ...(1)

I

n-1 n-2 n-3 -10 1 2 n-1 n

-1 n-1 n-2 n-30 1 2 n-1

n

0 = p A +p A +p A +...+p +p A1A =- [p A +p A +p A +...+p I]p

211121

112

tion)simplifica(on049λ6λλor

0λ211

1λ2111λ2

i.e.,0λIA

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Example 1:-

222121212221212222

211121

112

655565

556

655565

556

211121

112

211121

112

23

2

AAA

A

222121212221212222

655565

556

211121

112

100010001

0000000000

Now, pre – multiplying both sides of (1) by A-1 , we have

A2 – 6A +9I – 4 A-1 = 0 => 4 A-1 = A2 – 6 A +9I

311131113

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311131113

100010001

9211121

1126

655565

5564

1

1

A

A

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