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Suppose $(v_1 ... v_n )$ is linearly independent in a vector space V, and $w \in V$, if $(v_1 + w,.... v_n +w)$is linearly dependent, then $w \in span(v_1 ... v_n)$.

I'm still getting the hang of linear algebra proofs, which have a different "feel" to them than analysis proofs.

So, we know that $span (v_1 ... v_n ) = V$, so that it makes sense that $(v_1 + w,.... v_n +w)$ would be linearly dependent.

Because $w \in V$ , we know that $w$ must be able to be written as: $$w=\sum ^n _{i=1} a_i v_i$$

But how do you PROVE that? If we subtracted $w$ from $(v_1 + w,.... v_n +w)$, we would be left with the linearly independent list.

A nudge in the right directions is appreciated.

Astrum
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Hint: Supposing that $(v_1 + w, ..., v_n + w)$ is dependent, we may find constants $c_1, ..., c_n$ not all zero for which

$$c_1 (v_1 + w) + ... + c_n (v_n + w) = 0$$

Now rearranging a bit, we find

$$c_1 v_1 + ... + c_n v_n = - (c_1 + ... + c_n) w$$

Now consider two cases: $c_1 + ... + c_n = 0$, and not.

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    This is more than a nudge.. –  Aug 27 '13 at 07:34
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    @BryanUrízar Yes, apologies; I missed that point. I've pared down the response to (hopefully) something that's closer to what the asker would like. –  Aug 27 '13 at 07:36
  • If $c_1 + ... + c_n = 0$, that would mean that $c_1 v_1 + ... + c_n v_n = 0$, and hence would be independent.

    If $c_1 + ... + c_n $ is non-zero, it would therefore mean that $c_1 v_1 + ... + c_n v_n \neq 0$, and could not be independent.

     – Astrum Aug 27 '13 at 07:43
  • @Astrum Close, but not quite. If $c_1 v_1 + ... + c_n v_n = 0$, this contradicts that the vectors $v_1, ..., v_n$ are independent. –  Aug 27 '13 at 07:44
  • if all $c_n$ are zero, doesn't that prove linear independence? If not all $c_n$ are zero, that would mean dependence, and contradict the original claim that $(v_1 + ... + v_n)$ was an independent list. – Astrum Aug 27 '13 at 07:46
  • @Astrum By my first assumption, the $c_i$ are specifically not all zero. –  Aug 27 '13 at 07:48
  • Ah, ok, I see.

    So that would mean $\sum ^n _{i=1} c_i v_i = 0$, which means that it is dependent, and if it did not equal zero, we would have:

    $$-\frac{\sum ^n _{i=1} c_i v_i }{\sum ^n _{i=1} c_i} = w$$

     – Astrum Aug 27 '13 at 07:57
  • @Astrum Yep, you've got it. –  Aug 27 '13 at 07:59
  • And because the $c_n$ are just constants or arbitrary value, we could just say that

    $$\sum ^n _{i=1} b_i v_i = w$$

    Cool! Thanks for the help

     – Astrum Aug 27 '13 at 08:01
  • @Astrum Yes, with the specification that $b_i := -c_i / (c_1 + ... + c_n)$. –  Aug 27 '13 at 08:04