The Row Space If A is an m x n matrix, each row of A has n entries and thus can be identified with a...
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Transcript of The Row Space If A is an m x n matrix, each row of A has n entries and thus can be identified with a...
The Row SpaceIf A is an m x n matrix, each row of A has n entries and thus can be identified with a vector in Rn. The set of all linear combinations of the row vectors is called the row space of A and is denoted by Row A. …
Each row has n entries, so Row A is a subspace of Rn. Since the rows of A are identified with the columns of AT, we could also write Col AT in place of Row A.
-2 -5 8 0 -17
1 3 -5 1 5
3 11 -19 7 1
1 7 -13 5 -3
=
A
= (-2,-5,8,0,-17)
= (1,3,-5,1,5)
= (3,11,-19,7,1)
= (1,7,-13,5,-3)
1
2
3
4
r
r
r
r
The row space of A is the subspace of R5 spanned by {r1, r2, r3, r4 }.
If two matrices A and B are row equivalent, then their row spaces are the same. If B is in echelon form, the nonzero rows of B form a basis for the row space of A as well as B.
If A and B are row equivalentmatrices, then(a) A given set of columnvectors of A is linearlyindependent if and only if thecorresponding column vectorsof B are linearly independent.
(b) A given set of columnvector of A forms a basis forthe column space of A if andonly if the correspondingcolumn vector of B forms abasis for the column spaceof B.
Find the bases for the row and column spaces of
1 -3 4 -2 5 4
2 -6 9 -1 8 2
2 -6 9 -1 9 7
-1 3 -4 2 -5 -4
=
A
Solution Since elementary row operations do not change the row space of a matrix, we can find a basis for the row space of A by finding a basis for the row space of any row-echelon form of A.
1 -3 4 - 2 5 4
2 - 6 9 -1 8 2
2 - 6 9 -1 9 7
-1 3 -4 2 -5 - 4
1 -3 4 - 2 5 4
0 0 1 3 - 2 - 6
0 0 1 3 -1 -1
0 0 0 0 0 0
1 2
1 3
1 4
-2 R + R
-2 R + R
R + R
1 -3 4 - 2 5 4
0 0 1 3 - 2 - 6
0 0 0 0 1 5
0 0 0 0 0 0
2 3-1R + R
1 -3 4 -2 5 4
0 0 1 3 -2 -6
0 0 0 0 1 5
0 0 0 0 0 0
=
R
Now using the theorem (1) the non-zero vectors of R form a basis for the row space of R, and hence form bases for the row space of A. these bases vectors are
1 -3 4 -2 5 4
0 0 1 3 -2 -6
0 0 0 0 1 5
=
=
=
1
2
3
r
r
r
1 -3 4 -2 5 4
0 0 1 3 -2 -6
0 0 0 0 1 5
=
=
=
1
2
3
r
r
r
1 -3 4 -2 5 4
0 0 1 3 -2 -6
0 0 0 0 1 5
=
=
=
1
2
3
r
r
r
1 -3 4 -2 5 4
0 0 1 3 -2 -6
0 0 0 0 1 5
=
=
=
1
2
3
r
r
r
1 -3 4 -2 5 4
0 0 1 3 -2 -6
0 0 0 0 1 5
=
=
=
1
2
3
r
r
r
1 -3 4 -2 5 4
0 0 1 3 -2 -6
0 0 0 0 1 5
=
=
=
1
2
3
r
r
r
1 -3 4 -2 5 4
0 0 1 3 -2 -6
0 0 0 0 1 5
=
=
=
1
2
3
r
r
r
1 -3 4 -2 5 4
0 0 1 3 -2 -6
0 0 0 0 1 5
=
=
=
1
2
3
r
r
r
1 -3 4 -2 5 4
0 0 1 3 -2 -6
0 0 0 0 1 5
=
=
=
1
2
3
r
r
r
1 -3 4 -2 5 4
0 0 1 3 -2 -6
0 0 0 0 1 5
=
=
=
1
2
3
r
r
r
Keeping in mind that A and R may have different column spaces, we cannot find a basis for the column space of A directly from the column vectors of R. however, it follows from the theorem (2b) if we can find a set of column vectors of R that forms a basis for the column space of R, then the corresponding column vectors of A will form a basis for the column space of A.
The first, third, and fifth columns of R contains the leading 1’s of the row vectors, so
form a basis for the column space of R, thus the corresponding column vectors of A
1 4 5
0 1 -2
0 0 1
0 0 0
1 = = =
3 5c c c
1 4 5
0 1 -2
0 0 1
0 0 0
1 = = =
3 5c c c
form a basis for the column space of A.
1 4 5
2 9 8
2 9 9
-1 -4 -5
= = =
1 3 5c c c
Find bases for the space spanned by the vectors
= (1,-2,0,0,3) = (2,-5,-3,-2,6)
= (0,5,15,10,0) = (2,6,18,8,6)1 2
3 4
v v
v v
Solution Except for a variation in notation, the space spanned by these vectors is the row space of the matrix
1 -2 0 0 3
2 -5 -3 -2 6
0 5 15 10 0
2 6 18 8 6
Transforming Matrix to Row Echelon Form:
1 - 2 0 0 3
2 -5 -3 - 2 6
0 5 15 10 0
2 6 18 8 6
1 - 2 0 0 3
0 1 3 2 0
0 5 15 10 0
0 10 18 8 0
1 2
1 4
2
(-2)R + R
(-2)R + R
(-1)R
1 - 2 0 0 3
0 1 3 2 0
0 0 0 0 0
0 0 -12 -12 0
2 3
2 4
(-5)R + R
(-10)R + R
1 - 2 0 0 3
0 1 3 2 0
0 0 -12 -12 0
0 0 0 0 0
34R
1 - 2 0 0 3
0 1 3 2 0
0 0 1 1 0
0 0 0 0 0
3(-1/12)R
Therefore,
1 -2 0 0 3
0 1 3 2 0
0 0 1 1 0
0 0 0 0 0
=
R
The non-zero row vectors in this matrix are
These vectors form a basis for the row space and consequently form a basis for the subspace of R5 spanned by v1, v2, v3.
= (1,-2,0,0,3), = (0,1,3,2,0), = (0,0,1,1,0)1 2 3w w w
= (1,-2,0,0,3), = (0,1,3,2,0), = (0,0,1,1,0)1 2 3w w w
Find a basis for the row space of A consisting entirely of row vectors from A, where 1 -2 0 0 3
2 -5 -3 -2 6
0 5 15 10 0
2 6 18 8 6
=
A
Solution We will transpose A, thereby, converting the row space of A into the column space of AT; then we will use the method of example (2) to find a basis for the column space of AT; and then we will transpose again to convert column vectors back to row vectors. Transposing A yields
Transforming Matrix to Row Echelon Form:
1 2 0 2
-2 -5 5 6
0 -3 15 18
0 -2 10 8
3 6 0 6
=
TA
1 2 0 2
-2 -5 5 6
0 -3 15 18
0 - 2 10 8
3 6 0 6
1 2 0 2
0 -1 5 10
0 -3 15 18
0 - 2 10 8
0 0 0 0
1 2
1 5
2 R + R
(-3)R + R
1 2 0 2
0 1 -5 -10
0 -3 15 18
0 - 2 10 8
0 0 0 0
2(-1)R
1 2 0 2
0 1 -5 -10
0 0 0 -12
0 0 0 -12
0 0 0 0
2 3
2 4
(3)R + R
(2)R + R
1 2 0 2
0 1 -5 -10
0 0 0 1
0 0 0 -12
0 0 0 0
3(-1/12)R
1 2 0 2
0 1 -5 -10
0 0 0 1
0 0 0 0
0 0 0 0
3 412 R + R
1 2 0 2
0 1 -5 -10
0 0 0 1
0 0 0 0
0 0 0 0
=
R
The first, second and fourth columns contain the leading 1’s, so the corresponding column vectors in AT form a basis for the column space of AT; these are
Transposing again and adjusting the notation appropriately yields the basis vectors
for the row space of A.
1 2 2
-2 -5 6
and0 -3 18
0 -2 8
3 6 6
= , = =
1 2 4c c c
1 -2 0 0 3 2 -5 -3 -2 6 2 6 18 8 6= , = and =1 2 4r r r
The main result of this lecture involves the three spaces: Row A, Col A, and Nul A. The following example prepares the way for this result and shows how one sequence of row operations on A leads to bases for all three spaces.
Find bases for the row space, the column space and the null space of the matrix
-2 -5 8 0 -17
1 3 -5 1 5
3 11 -19 7 1
1 7 -13 5 -3
=
A
The rank of A is the dimension of the column space of A. Since Row A is the same as Col AT, the dimension of the row space of A is the rank of AT. The dimension of the null space is sometimes called the nullity of A.
The dimensions of the column space and the row space of an m x n matrix A are equal. This common dimension, the rank of A, also equals the number of pivot positions in A and satisfies the equation
rank A + dim Nul A = n
The Rank Theorem
If A is an m x n, matrix, then(a) rank (A) = the number of
leading variables in the solution of Ax = 0
(b) nullity (A) = the number of parameters in the general solution of Ax = 0
Find the number of parameters in the solution set of Ax = 0 if A is a 5 x 7 matrix of rank 3.
nullity (A)=n-rank (A)=7-3=4There are four parameters.
If A is any matrix, then
rank (A) = rank (AT)
Four Fundamental Matrix Spaces
Row space of A
Column space of A
Null space of A
Null space of AT
Suppose that A is an m x nmatrix of rank r, AT is an n x mmatrix of rank r .Fundamental space
DimensionRow space of A rColumn space of A rNull space of A n-rNull space of AT m-r
If A is a 7 x 4 matrix, then the rank of A is at most 4 and, consequently, the seven row vectors must be linearly dependent. …
If A is a 4 x 7 matrix, then again the rank of A is at most 4 and, consequently, the seven column vectors must be linearly dependent.
Let A be an n x n matrix. Then the following statements are each equivalent to the statement that A is an invertible matrix. …
The columns of A form a basis of Rn.
Col A = Rn
dim Col A = n
rank A = n
Nul A = {0}
dim Nul A = 0
The matrices below are row
equivalent2 -1 1 -6 8
1 -2 -4 3 -2
-7 8 10 3 -10
4 -5 -7 0 4
=
A
1 -2 -4 3 -2
0 3 9 -12 12
0 0 0 0 0
0 0 0 0 0
=
B
1. Find rank A and dim Nul A.2. Find bases for Col A andRow A.3. What is the next step toperform if one wants to find abasis for Nul A?4. How many pivot columnsare in a row echelon form ofAT?
