How to prove stuff about linear algebra?

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To prove that if (v_1 + w, v_2 + w, ..., v_n + w) is linearly dependent, then w must be in the span of (v_1, ..., v_n), one starts with the assumption of linear independence of (v_1, v_2, ..., v_n). The proof shows that the linear dependence of the modified vectors leads to a contradiction unless w can be expressed as a linear combination of the original vectors. The discussion highlights a common misunderstanding regarding the definition of linear independence, clarifying that the coefficients must be zero for the equation to hold true. Overall, the proof is largely correct, but requires precise language to avoid confusion. Clear communication of mathematical concepts is essential for effective understanding in linear algebra.
*melinda*
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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!
 
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Your proof is pretty much correct, but in this sentence:
*melinda* said:
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
 
Yes, that would make a bit more sense. Sometimes I understand what I mean to do, but don't know how to say it. :rolleyes:

Thanks a bunch!
 
*melinda* said:
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.
 
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