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The theorem states that if $(v_1, ...,v_m)$ is linearly dependent in $V$ and $v_1 \neq 0$ then there exists $j \in \{2,...,m\}$ such that $v_j \in span(v_1,...,v_{j-1})$.

If $(v_1, v_2, ..., v_m)$ are linearly dependent, then by definition of linear dependence, at least one of these vectors can be expressed as a linear combination of remaining vectors.

To be more precise, vectors are linearly dependent if not all scalars $a_i$ have to be zero in this equality:

$$a_1 v_1 + a_2 v_2 + ... + a_n v_n = 0$$

For instance if $0v_1+0v_2+3v_3=0$, then $v_2=-3v_3$. So you just pick the vector $v_j$ that's associated with non-zero $a_j$, subtract everything else from both sides and get $v_j$ expressed as a linear combination of remaining vectors. Therefore $v_j \in span(v_1,...,v_{j-1})$, by definition of span.

I just don't understand the requirement that $v_1 \ne 0$. It always works, no matter if $v_1$ is zero or not.

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The requirement that $v_1 \neq 0$ is necessary. For instance, let $V = \mathbb{R}^2$, $v_1 = 0$ and $v_2 = \textbf{i}$, which form a linearly dependent set.

Then the only $j$ you can pick is $2$, but $v_2 \notin <v_1>$.

Furthermore, note that your proof as written is not correct. You have to express the $v_j$ as a linear combination of $v_i$s where $i< j$. But, there is no reason to suppose that you don't use $v_i$s with $i>j$ with your construction.

I'll show one way to do the argument correctly.

Let $V$ be a vector space, $v_1,\ldots,v_m\in V$ linearly dependent with $v_1 = 0$. Then by linear dependence there exists scalars $c_1,\ldots,c_k$ not all $0$ so that $$c_1v_1 + \ldots + c_mv_m = 0$$

Let $i\in\{1,\ldots,m\}$ be the greatest index so that $c_i \neq 0$. If $i=1$ then we have $$ c_1 v_1 = 0$$ and so, $v_1 = 0$ contradicting our hypothesis. Hence $i > 1$, and therefore $$ c_1 v_1 + \cdots + c_i v_i = 0 $$ and thus $c_iv_i = -(c_1v_1 + \cdots + c_{i-1} v_{i-1})$. By definition of $i$, $c_i \neq 0$ and so $$v_i = \frac{-1}{c_i}(c_1v_1 + \cdots + c_{i-1}v_{i-1})$$ Consequently $v_i \in \text{span}(v_1,\ldots,v_{i-1})$ and so $i$ an example of the required $j$.

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  • $\begingroup$ 'If $i=1$ then we have $ c_1 v_1 = 0$ and so, $v_1 = 0$ contradicting our hypothesis.' - I don't see any contradiction here. Could you explain? $\endgroup$ – user4205580 Jan 16 '16 at 16:55
  • $\begingroup$ We assumed $v_1\neq 0$ $\endgroup$ – James Jan 16 '16 at 16:59
  • $\begingroup$ Okay, but the point was to show that assumption $v_1 \ne 0$ is necessary. Is it necessary because $c_1 v_1 = 0$ implies that $v_1$ cannot be expressed as a linear combination of vectors whose index is less than $1$, simply because there are no such vectors (and the theorem says every such vector CAN be expressed that way)? By assuming $v_1 \ne 0$ we eliminate this problem - I think I get it (or not)? $\endgroup$ – user4205580 Jan 16 '16 at 17:02
  • $\begingroup$ No, you don't get it. There are two issues, the first is that the proof you give of the proposition is incorrect, this is addressed by everything after "I'll show one way to do the argument correctly." The second issue is whether or not you need the hypothesis that $v_1\neq 0$, I showed that you do in the first few paragraphs by giving an example of a situation which satisfies every hypothesis except "$v_1\neq 0$" and showed that the conclusion is not true, in the example I give. $\endgroup$ – James Jan 16 '16 at 17:09
  • $\begingroup$ I think you are confusing a few issues. I used the hypothesis that $v_1 \neq 0$ in my proof of the proposition. This does not mean that the theorem needs this hypothesis. To show that a hypothesis is necessary for a theorem to hold requires you to give a counter-example to the theorem when the hypothesis doesn't hold. $\endgroup$ – James Jan 16 '16 at 17:22
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A $0$ vector can never be part of a linearly independent list of vectors. In other words, a list of vectors containing a $0$ vector is always linearly dependent.

Therefore consider a list of vectors $(v_2, v_3, ... , v_n)$ which are linearly independent. Now $(0, v_2, v_3, ... , v_n)$ is linearly dependent, however there exists no $j \in \{2,...,n\}$ such that $v_j \in span(0,v_2,...,v_{j-1})$ by the choice of the list $(v_2, v_3, ... , v_n)$.

I guess you are mostly uncomfortable with the special treatment of $v_1$. The lemma could've been stated as for a list of linearly dependent vectors, there exists at least one vector which belongs to the span of the rest of the vectors and in that case there is no need of the special treatment of the first vector. This is because the $0$ vector can always be expressed as a linear combination of other vectors (a linear combination where all the coefficients are $0 \in \mathbb{F}$). However there are a lot to be gained by this particular way of stating the lemma as it makes some other proofs really easy and elegant.

Some books do not pose any restriction on $v_1$ and just states that $\exists j$ such that $v_j \in span(v_1, ..., v_{j-1})$. Now this would mean if $j=1$, then $v_1 \in span()$. In this case they define $span()$ as the empty vector space or $\{0\}$. So naturally $v_1 \in span() \implies v_1 = 0$.

Hope this answers your question.

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