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$\textbf{Definition}.$

Let $V$ be a vector space, and let $\textbf{v}_1,\dots,\textbf{v}_n \in V$. Let $\alpha_1,\dots,\alpha_n$ be scalars. Let $\textbf{0}$ be the zero element of $V$.

$\textbf{v}_1,\dots,\textbf{v}_n$ are said to be linearly independent if $$\alpha_1 \textbf{v}_1 + \dots + \alpha_n \textbf{v}_n = \textbf{0} \Leftrightarrow \alpha_1,\dots,\alpha_n = 0$$

Firstly, is this a valid definition of linear independence?

Secondly, how do I find the negation of this definition of linear independence? I would expect to get something like there exist scalars $\alpha_1,\dots,\alpha_n$, not all zero, such that $\alpha_1 \textbf{v}_1 + \dots + \alpha_n \textbf{v}_n = \textbf{0}$, but I am not sure how I would arrive at something like this.

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    $\begingroup$ You are right about the negation. $\endgroup$ – xbh Jul 31 '18 at 5:19
  • $\begingroup$ @xbh Yes, I suspected so, but I would like to know how to (i.e go through some explicit steps) to negate the definition to arrive at something equivalent to what I wrote at the end. $\endgroup$ – user445909 Jul 31 '18 at 5:21
  • $\begingroup$ Reverse the quantifiers, i.e. "for all" to "exist one", "exist one" to "for all". Also take the negation of the statement. $\endgroup$ – xbh Jul 31 '18 at 5:23
  • $\begingroup$ Example. The definition of linear independence could be rewritten as "if $\sum \alpha_j \boldsymbol v_j = \mathbf 0, $then all $\alpha_j = 0$" [the other direction holds always under the context of vector space]. Now take the negation of the conclusion above: "all" becomes "some", "$\alpha_j = 0$" becomes "$\alpha_j \neq 0$". Combine them together yields "some $\alpha_j $can be nonzero despite that equation holds". $\endgroup$ – xbh Jul 31 '18 at 5:30
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    $\begingroup$ @J.G. Thanks for pointing out. I have noticed these flaws. $\endgroup$ – xbh Jul 31 '18 at 5:34
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Your definition of linear dependence is valid, although we usually only write $\implies$ since the left-hand arrow is trivial. Of course, to say we can deduce all $\alpha_i$ is equivalent to saying there does not exist any other choice of the $\alpha_i$ that works. Therefore, the negation is as expected to say that one does.

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    $\begingroup$ I guess this is pretty obvious, thank you. I was wishing to see for myself that what you have said follows from explicit negation of the definition using predicate calculus, but I have now convinced myself of this by writing it out on paper. $\endgroup$ – user445909 Jul 31 '18 at 5:32

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