How to prove stuff about linear algebra?

[*melinda*]

How to prove stuff about linear algebra?

Question:

Suppose $(v_1, v_2, ..., v_n)$ is linearly independent in $V$ and $w\in V$.
Prove that if $(v_1 +w, v_2 +w, ..., v_n +w)$ is linearly dependent, then $w\in span(v_1, ...,v_n)$.

To prove this I tried...

If $(v_1, v_2, ..., v_n)$ is linearly independent then $a_1 v_1 + ...+a_n v_n =0$ for all $(a_1 , ..., a_n )=0$.
then,
$a_1 (v_1 +w)+a_2 (v_2 +w)+...+a_n (v_n +w)=0$
is not linearly independent, but can be rewritten as,
$a_1 v_1 + ...+a_n v_n +(\sum a_i )w=0$
so,
$a_1 v_1 + ...+a_n v_n = -(\sum a_i )w$.
Since $w$ is a linear combination of vectors in $V$, $w\in span(V)$.

Did I do this right?
Is there a better way of doing this?
Any input is much appreciated!

 

[LeonhardEuler]

Your proof is pretty much correct, but in this sentence:

If $(v_1, v_2, ..., v_n)$ is linearly independent then $a_1 v_1 + ...+a_n v_n =0$ for all $(a_1 , ..., a_n )=0$.

I think you mean to say:
If $(v_1, v_2, ..., v_n)$ is linearly independent then $a_1 v_1 + ...+a_n v_n =0$ only when each $a_i=0$

 

[*melinda*]

Yes, that would make a bit more sense. Sometimes I understand what I mean to do, but don't know how to say it.

Thanks a bunch!

 

[Hurkyl]

If $(v_1, v_2, ..., v_n)$ is linearly independent then $a_1 v_1 + ...+a_n v_n =0$ for all $(a_1 , ..., a_n )=0$.

This is wrong. If the collection of vectors is independent, and if $a_1 v_1 + ...+a_n v_n =0$ then $a_1 = a_2 = \cdots = 0$.