# Linear Algebra - Linear Independence and Span

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.

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.

• This is more than a nudge.. – user70962 Aug 27 '13 at 7:34
• @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. – user61527 Aug 27 '13 at 7:36
• 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 7:43
• @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. – user61527 Aug 27 '13 at 7:44
• 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 7:46