Question

A=[ 7 8 -2 -6 7 4 1 ; 2 4 -4 -13 9 9 -12 ; 6 6 0 -9 8 9 -4 ; 1 8 -14 -22 5 8 -1 ; 4 9 -10 -14 7 4 -1]

B=[ 19 4 4 14 -3 -7 -5 ; 21 -6 -5 10 14 -2 4 ; 22 -4 5 13 5 -6 4 ; 41 20 0 26 11 -1 -27 ; 29 14 -2 20 3 -4 -19]

Remark: You may only use rref , or commands that are covered in our MATLAB guides, and no other special MATLAB functions for this problem.

(a) (3 points) Find a basis for Col(A) and determine its dimension.

(b) (5 points) Find a basis for Nul(A) and determine its dimension.

(c) (3 points) Find a basis for Row(A) and determine its dimension.

(d) (5 points) Find a basis for the left nullspace of A and determine its dimension.

(e) (2 points) Find a basis for Col(B) and determine its dimension.

(f) (6 points) Determine whether Col(A) = Col(B), i.e. determine whether the two given subspaces of R 5 are the same or not. If not, find a vector v

that belongs in one of the subspaces and not the other.

(g) (6 points) Determine whether Nul(A) = Nul(B). If not, find a vector v that belongs in one of the subspaces and not the other.

Answer 1

A_r=[1 0 2 0 0 2 1
0 1 -2 0 0 -2 1
0 0 0 1 0 -1 0
0 0 0 0 1 0 -2
0 0 0 0 0 0 0] ----------------->rref of A

To obtain basis for column space we use pivot columns from the Matrix A

so column_basis(A)=

[{7 { 8 {7 { 8
2 4 2 4
6 6 6 6
1 8 1 8
4 } 9 } 4 } 9}]

Dimension is Rank4

b)A_r=[1 0 2 0 0 2 1
0 1 -2 0 0 -2 1
0 0 0 1 0 -1 0
0 0 0 0 1 0 -2
0 0 0 0 0 0 0] ----------------->rref of A

A_r=[1 0 2 0 0 2 1 0
0 1 -2 0 0 -2 1 0
0 0 0 1 0 -1 0 0
0 0 0 0 1 0 -2 0
0 0 0 0 0 0 0 0]

Again back to equivalent system:

x1+0+2x3+0+0+2x6+x7=0

0+x2-2x3+0+0-2x6+x7=0

x3=x3%add free varaiabe in the equation

0+0+0+x4+0-x6+0=0

0+0+0+0+x5+0-2x7=0

x7=x7 %free variable

on simplification:

x1=-2x3-2x6-x7

x2=2x3+2x6-x7

x3=x3

x4=x6

x5=2x7

x6=x6

x7=x7

Arranging the terms we get

[x1 [-2x3 [-2x6 [-x7

x2 2x3 2x6 x7

x3 x3 0 0

x4 = 0 x6 0

x5 0 0 2x7

x6 0 x6 0

x7] 0] 0] x7]

basis of null sapce

[2 -2 -1

2 2 -1

1 0 0

0 1 0

0 0 2

0 1 0

0 0 1]

dimension=Rank3

c)Basis of Row

c)A_r=[1 0 2 0 0 2 1
0 1 -2 0 0 -2 1
0 0 0 1 0 -1 0
0 0 0 0 1 0 -2
0 0 0 0 0 0 0] ----------------->rref of A

Non zero rows in RREF is row basis

Basis of Row(A)={[1 0 2 0 0 2 1]

[ 0 1 -2 0 0 -2 1]

[ 0 0 0 1 0 -1 0]

[ 0 0 0 0 1 0 -2]}

Dimension=Rank4

e)

Col space of B

B_r=[ 1.0000 0 0 0 2.2000 1.0000 -0.2000
0 1.0000 0 0 0.2000 0.5000 -1.2000
0 0 1.0000 0 -0.2000 0 0.2000
0 0 0 1.0000 -3.2000 -2.0000 0.2000
0 0 0 0 0 0 0]----rref(B)

basis of column B

[{ 19 {4 {4 {14
21 -6 -5 10
22 -4 5 13
41 20 0 26
29} 14} -2} 20}]

Dimension is Rank4

d)left null space :Nullspace (AT)[A transpose]

rreft(AT)=[ 1.0000 0 0 0 1.1363
0 1.0000 0 0 0.4340
0 0 1.0000 0 -0.8750
0 0 0 1.0000 0.4279
0 0 0 0 0
0 0 0 0 0
0 0 0 0 0]

after apply Null space procedure as in b

gives the basis of left null space as

{[-1309/1152

-125/288

7/8

-493/1152

1]}

Dimesnion:Rank1