For each of the following subspaces S of the given vector sp

For each of the following subspaces S of the given vector spaces V ,

find a subset L that is independent and has the same linear span.
1 2 1 2 1 0 0 1

2. For each of the following subspaces S of the given vector spaces V, find a subset L that is independent and has the same linear span. 1 2 2 1 01 0 1 11 (a) Let V M2 (R), and S 3 4 5 0 -1 2 2 -1 (b) Let P (R), and S {r2 r 2, z -1, z 2r

Solution

(a) Let the standard basis for M₂® be denoted by {e₁, e₂, e ₃, e₄} . Then the matrices in S equal A₁ = e₁+2e₂ +3e₃ +4e₄, A₂ = -e₁ +2e₂ +5e₃ , A₃ = e₁-e₃+2e₄ , and A₄ = e₂ +2e₃ –e₄ . Further, let A =

1

-1

1

0

2

2

0

1

3

5

-1

2

4

0

2

-1

We will reduce A to its RREF as under:

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

Add -4 times the 1st row to the 4th row ; Multiply the 2nd row by ¼

Add -8 times the 2nd row to the 3rd row ; Add -4 times the 2nd row to the 4th row

Interchange the 3rd row and the 4th row ; Multiply the 3rd row by -1/2

Add -1/4 times the 3rd row to the 2nd row ; Add 1 times the 2nd row to the 1st row

Then the RREF of A is

1

0

1/2

0

0

1

-1/2

0

0

0

1

0

0

0

0

1

Apparently, A₃= 1/2A₁-1/2A₂ . Thus, L = { A₁, A₂, A₄} , where A₁ =

1

2

3

4

A₂ =

-1

2

5

0

and A₄ =

0

1

2

-1

(b) Let A =

1

0

1

-1

1

-2

-2

-1

3

We will reduce A to its RREF as under:

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

Add 1 times the 2nd row to the 3rd row; Multiply the 3rd row by ¼

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

Then the RREF of A is

1

0

0

0

1

0

0

0

1

This implies that the vectors in S are linearly independent so that L = S.

1	-1	1	0
2	2	0	1
3	5	-1	2
4	0	2	-1