Prove that a set is a basis...

**whoadude16** · July 12th 2010, 06:38 PM

Let {v1,v2,v3} be a basis for vector space V. Prove that, if w is not in sp(v1,v2), then S = {v1,v2,w} is also a basis for V.

I know that in order for S = {v1,v2,w} to be a basis for V, the set S must span V and S must be linearly independent. However, I have no idea whatsoever on how to start this proof. Please help.

**roninpro** · July 12th 2010, 06:50 PM

You've identified the necessary items. We can start from there.

First, suppose that $S=\{v_1, v_2, w\}$ is linearly dependent. Then, we can find $c_1,c_2,c_3$ (not all zero) so that $c_1v_1+c_2v_2+c_3w=0$ , and we may assume that $c_3\ne 0$ . (Why?) Rearranging gives $w=\frac{c_1v_1+c_2v_2}{c_3}$ . Any problems with this?

To show $\text{span } S=V$ , we can write $w=b_1v_1+b_2v_2+b_3v_3$ for some $b_1,b_2,b_3$ , where, in particular, $b_3\ne 0$ . Now, $S=\{v_1, v_2, b_1v_1+b_2v_2+b_3v_3\}$ . Can you see how this helps?

Good luck.

**whoadude16** · July 12th 2010, 07:29 PM

Originally Posted by roninpro

You've identified the necessary items. We can start from there.

First, suppose that $S=\{v_1, v_2, w\}$ is linearly dependent. Then, we can find $c_1,c_2,c_3$ (not all zero) so that $c_1v_1+c_2v_2+c_3w=0$ , and we may assume that $c_3\ne 0$ . (Why?) Rearranging gives $w=\frac{c_1v_1+c_2v_2}{c_3}$ . Any problems with this?

To show $\span S=V$ , we can write $w=b_1v_1+b_2v_2+b_3v_3$ for some $b_1,b_2,b_3$ , where, in particular, $b_3\ne 0$ . Now, $S=\{v_1, v_2, b_1v_1+b_2v_2+b_3v_3\}$ . Can you see how this helps?

Good luck.

Okay here goes:
$w=\frac{c_1v_1+c_2v_2}{c_3}$ this basically states that w is dependent of (v1,v2) but since it was given that w is not in span(v1,v2), we can say that c1=c2=c3=0 and S is linearly independent ?

Span(S) = {rv1 + sv2 + t( $b_1v_1+b_2v_2+b_3v_3$ )} = {c1v1 + c2v2 + c3v3} = Span(V) = V ?

Is that right?...

**roninpro** · July 12th 2010, 08:57 PM

You are correct with linear independence. However, $\text{span } S=V$ doesn't quite follow from what you have written.

We do have $\text{span } S=\{rv_1+sv_2+t(b_1v_1+b_2v_2+b_3v_3)\ | \ r,s,t\in F\}$ , as you wrote. But this can be rewritten: $\{(r+tb_1)v_1+(s+tb_2)v_2+tb_3v_3\ | \ r,s,t\in F\}$ . To complete the proof, you need to show that $r+tb_1, s+tb_2, tb_3$ range over all possible values, which amounts to solving a system of equations.

**whoadude16** · July 13th 2010, 07:03 PM

Originally Posted by roninpro

You are correct with linear independence. However, $\text{span } S=V$ doesn't quite follow from what you have written.

We do have $\text{span } S=\{rv_1+sv_2+t(b_1v_1+b_2v_2+b_3v_3)\ | \ r,s,t\in F\}$ , as you wrote. But this can be rewritten: $\{(r+tb_1)v_1+(s+tb_2)v_2+tb_3v_3\ | \ r,s,t\in F\}$ . To complete the proof, you need to show that $r+tb_1, s+tb_2, tb_3$ range over all possible values, which amounts to solving a system of equations.

Okay... Thanks alot for your help !!
Also, i just learned in class today that since sp(v1,v2,v3) is a basis for vector space V, we know that dim(V)=3 - which means that any subset of vectors with 3 elements is also a basis for V. I hope that's right.
Thanks again for your help !!

**Ackbeet** · July 13th 2010, 07:05 PM

"...we know that dim(V)=3 - which means that any subset of vectors with 3 elements is also a basis for V..."

They have to be linearly independent. Then they are a basis.

**whoadude16** · July 13th 2010, 07:59 PM

Originally Posted by Ackbeet

"...we know that dim(V)=3 - which means that any subset of vectors with 3 elements is also a basis for V..."

They have to be linearly independent. Then they are a basis.

...Yea so i dont have to show that span(S) = V...

**whoadude16** · July 13th 2010, 08:01 PM

BTW... thank you all for your help.. this forum is GREAT !!!

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