Let S 2 1 1 3 1 1 2 3 6 3 3 9 1 1 2 2 2 2 6 3 Find a basis

Let S = {[-2 -1 -1 3], [-1 1 2 3], [6 3 3 -9], [1 1 -2 -2], [2 -2 6 -3]}. Find a basis for the subspace W spanned by S, and the dimension of W.

Solution

Let A =

-2

-1

6

1

2

-1

1

3

1

-2

-1

2

3

-2

6

3

3

-9

-2

-3

We will reduce A to its RREF as under:

Multiply the 1st row by -1/2

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

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

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

Multiply the 2nd row by 2/3

Add -5/2 times the 2nd row to the 3rd row

Add -3/2 times the 2nd row to the 4th row

Multiply the 3rd row by -3/10

Add 1 times the 3rd row to the 4th row

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

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

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

Then the RREF of A is

1

0

-3

0

-2

0

1

0

0

-1

0

0

0

1

-3

0

0

0

0

0

It may now be observed that (6,3,3,-9)^T = -3(-2,-1,-1,3)^T and (2,-2,6,-3) = -2(-2,-1,-1,3)^T-1(-1,1,2,3)^T- 3(1,1,-2,-2)^T. The other 3 vectors are linearly independent. Hence a basis for W is {(-2,-1,-1,3)^T,(-1,1,2,3)^T, (1,1,-2,-2)^T}. Further, dim(W) = 3.

-2	-1	6	1	2
-1	1	3	1	-2
-1	2	3	-2	6
3	3	-9	-2	-3