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I'm trying to interpret the very last sentence. He says nothing about $a_{1}$, so in the case of $a_{1}=0$ all the other coefficients can not be zero, which would imply linearly independence. The case where $a_{1}\neq 0$ all the other coefficients can not be zero because, if they were, we would have $a_{1}v_{1}=0$ which is never true since $v_{1}\neq 0$.

Is this correct? Thank you in advance.

I have seen this post post about lin.dep. lemma., but I think my question is unanswered in that post.

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Community
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New_to_this
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3 Answers3

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What you've said is perfectly correct. A slightly quicker/neater/different way to see it (without splitting into cases) might be the following:

Suppose instead that all of $a_2,\dots,a_m$ are equal to zero. Then $a_1v_1=0$. But $v_1\not=0$ so it must be that $a_1=0$, contradicting our original assumption that not all of the $a_i$ are $0$.

Tom Oldfield
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It seems to combine two cases together:

  • If $a_0=0$, then since $\{a_1,a_2,\ldots,a_m\}$ contains a non-zero element and $a_0=0$, there must be a non-zero member of $\{a_2,a_3,\ldots,a_m\}$.

  • If $a_0 \neq 0$, then $a_1 v_1 \neq 0$ since $v_1 \neq 0$. Hence, if $a_1 v_1 + x=0$ where $x=a_2v_2+a_3v_3+\cdots+a_mv_n$, then $x = -a_1v_1 \neq 0$ and hence $\{a_2,a_3,\ldots,a_m\}$ cannot all be zero.

Rebecca J. Stones
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If you note that $span{} = \{0\}$, then it is clear that $v_1$ can also be equal to $0$, because if we have an empty span generated by an empty set of preceding vectors, then if $v_1 \in span\{0\}$, $v_1 = 0$.

The empty set is independent by itself but when an extra $v_1 = 0$ gets added, then the set of vectors becomes linearly dependent.

Stating that $v_1 \neq 0$ is kind of a confusing way to prove this theorem IMO.

St Vincent
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  • If $v_{1}=0$ then none of the other scalars had to be zero. This can not happen, since in the next step of the prove, the author divides by $a_{j}$ where $j\in{2,\ldots,m}$. – New_to_this Sep 19 '13 at 15:57
  • Well, we know that $a_j \neq 0$, but this does not mean that $v_1 \neq 0$. The theorem is correct, but we can drop the case where $v_1 \neq 0$. $span(v_1,\ldots,v_{j-1}) = {0}$ for $j = 1$, which can be written as $v_1 \in span()$ which is just the empty list. The empty list by itself is independent but if $v_1$ is an element of $span()$ then $v_1 = 0$ which will be a linearly dependent vector inside the list. – St Vincent Sep 19 '13 at 19:27