Consider the 3 times 3 matrices A 1 1 1 0 1 0 1 0 1 B 1 1

Consider the 3 times 3 matrices A = (1 1 1 0 1 0 1 0 1), B = (1 1 1 1 0 1 1 1 1). Find of the image and kernel of A and B and decide whether Im(A) = Im(B) and whether ker(A) = ker(B).

Solution

We will reduce A to its RREF as under:

Add -1 times the 1st row to the 2nd row

Add -1 times the 1st row to the 3rd row

Then the RREF of A is

1

0

1

0

1

-1

0

0

0

Then Im (A) = Col(A). Its basis is { (1,0,0)^T, (0,1,0)^T}. The Kernel of A is the set of solutions to the equation AX = 0. If X = (x,y,z)^T, then it is equivalent to x+z= 0 and y-z = 0, or, x = -z and y = z so that X = ( -z,z,z)^T = z(-1,1,1)^T. Thus a basis for Ker (A) is {(-1,1,1)^T}.

We will reduce B to its RREF as under:

Add -1 times the 1st row to the 2nd row

Add -1 times the 1st row to the 3rd row

Multiply the 2nd row by -1

Add -1 times the 2nd row to the 1st row

Then the RREF of B is

1

0

1

0

1

0

0

0

0

Then Im (B) = Col(B). Its basis is { (1,0,0)^T, (0,1,0)^T}. The Kernel of B is the set of solutions to the equation BX = 0. If X = (x,y,z)^T, then it is equivalent to x+z= 0 and y = 0, so that x = -z . Hence X = ( -z,0,z)^T = z(-1,0,1)^T. Thus a basis for Ker (B) is {(-1,0,1)^T}.

Thus Im(A) = Im(B) but Ker(A) Ker(B)

1	0	1
0	1	-1
0	0	0