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Let $v_1, \cdots, v_n \in V$ where $V$ a vector space. Exhibit that the following are equivalent:

(1) The vectors are linearly dependent

(2) There exist scalars $a_1, \cdots, a_n$ not all zero such that $\sum_{i=1}^{n} a_iv_i = 0$

(3) For some $0 \le i < n$ we have $v_i \in \operatorname{span}\{v_j \ | \ 1 \le j < i\}$

(4) For some $0 \le i \le n$ we have $v_i \in \operatorname{span}\{v_j \ | \ 1 \le j \le n,\ \ i \neq j\}$

I am only left with proving that $(2) \implies (3) \implies (4)$. Any help?

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In (4), $0\le i\le n$ should probably read $0\le j\le n$. – user1551 Feb 20 '13 at 7:46
Absolutely! In (3) as well! – user44069 Feb 20 '13 at 7:48
Shouldn't it be $0 < j$? The lowest index in your family of vectors is 1, not 0. – Jack M Feb 20 '13 at 7:49
As well, on the left, the interval should be $0 < i \leq n$ as you number your vectors from 1-n. – gnometorule Feb 20 '13 at 7:49
@JackM Good eyes! – user1551 Feb 20 '13 at 7:50
up vote 4 down vote accepted

Just for 2 to 3: Let there are $a_1, a_2,...,a_n$ not all zero such that $\sum_{i=1}^n a_iv_i=0$. So at least we know there is an $a_k,~~1\leq k\leq n$ such that $a_k\neq 0$. W e have $a_1v_1+a_2v_2+...+a_kv_k+...a_nv_n=0$ and so $$v_k=\frac{a_1v_1+a_2v_2+...+a_{k-1}v_k+a_{k+1}v_{k+1}...a_nv_n}{-a_k}$$ This means 3 is held.

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I think this is incorrect. You are only allowed to have vectors that are of lower index than $k$ which is clearly not the case here. – user44069 Feb 20 '13 at 7:51
@Stefan: it's essentially correct. I added a comment to my answer below to finish this off. – gnometorule Feb 20 '13 at 7:56
Thanks a lot! This clarifies things! – user44069 Feb 20 '13 at 7:58
Good observations! +1 – amWhy Feb 20 '13 at 14:01
When do you sleep, Babak? ;-) – amWhy Feb 20 '13 at 14:17

(3) $\Rightarrow$ (4) is trivial: write the vector $v_i$ from (3) as a linear combination of those with indices $1, \dots, i-1$ (which is what (3) says you can do), and choose $a_j = 0$ for all (if any) $j > i$.

Edit: to finish the proof of (2) to (3) of the other answer, choose $i$ such that $a_i$ is the largest non-zero coefficient you get from (2). Then everything goes through as in the proof.

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Thanks for what you have done for clarify the main point. I usally assume some things are clear o the OP. Thanks again. – Babak S. Feb 20 '13 at 8:31
@BabakS.: NP! A little remark was all it took. – gnometorule Feb 20 '13 at 8:33

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