Let A 1 0 2 0 2 1 3 1 3 2 4 2 4 3 5 4 5 4 6 6 Find a basis

Let A = (1 0 -2 0 2 1 -3 1 3 2 -4 2 4 3 -5 4 5 4 -6 6). Find a basis for (a) the row space of A. (b) the column space of A. (c) the null space of A.

Solution

We will first reduce A to its RREF as under:

Add 2 times the 1st row to the 3rd row

Add -1 times the 2nd row to the 3rd row

Add -1 times the 2nd row to the 4th row

Interchange the 3rd row and the 4th row

Add -3 times the 3rd row to the 2nd row

Add -4 times the 3rd row to the 1st row

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

Then the RREF of A is

1

0

-1

0

1

0

1

2

0

-2

0

0

0

1

2

0

0

0

0

0

Then a basis for Col(A) is { (1,0,-2,0)^T, ( 2,1,-3,1)^T, (4,3,-5,4)^T}

The null space of A is the set of solutions to the equation AX = 0. If X = (x₁,x,x₃,x₄,x₅)^T, then this equation is equivalent to x–x₃+x₅=0, x₂+2x₃-2x₅=0 and x₄+2x₅=0. Let x₃=r and x₅=t. Then X = (r-t, -2r+2t, r, -2t, t)^T =

r( 1,-2,1,0,0)^T +t ( -1,2,0, -1,1)^T. Thus, a basis for Null(A) is { ( 1,-2,1,0,0)^T, ( -1,2,0, -1,1)^T}.

Similarly, the RREF of A^T is

1

0

-2

0

0

1

1

0

0

0

0

1

0

0

0

0

0

0

0

0

Then a basis for Row (A) is{ (1,2,3,4,5), (0,1,2,3,4),(0,1,2,4,6)}.

1	0	-1	0	1
0	1	2	0	-2
0	0	0	1	2
0	0	0	0	0