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If you have vector $v$ and vector $u$ that are linearly independent to each other. And vector $u$ is also linearly independent to vector $w$ and $w$ is not equal to vector $v$, then does that imply vector $v$ and vector $w$ are also linearly independent?

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    $\begingroup$ Incidentally, transitivity doesn't exclude the $w \neq v$ case. $\endgroup$
    – user14972
    Sep 3 '17 at 1:17
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No. If $u$ and $v$ are linearly independent, then $v$ and $cu$ are linearly independent (where $c$ is a scalar constant). But $u$ and $cu$ are not linearly independent, hence transitivity fails.

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    $\begingroup$ Curiously, in the case of a vector space over the $2$-member field $\{0,1\},$ linear independence, in the case $w\ne v,$ actually $is$ transitive.................+1 $\endgroup$ Sep 3 '17 at 3:27
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    $\begingroup$ @DanielWainfleet I like it. I'm going to have to add that to my collection of pathological examples. $\endgroup$
    – Xander Henderson
    Sep 3 '17 at 3:40
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No. In the plane, let $\mathbf{u} = \langle 1, 0 \rangle$, $\mathbf{v} = \langle 0, 1 \rangle$, and $\mathbf{w} = \langle 0, 2 \rangle$.

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As a general principle, reflexivity is necessary for transitivity, as you can always just compose $u \sim v$ with $v \sim u$, and if the relation was transitive, this would yield $u \sim u$. So if the relation is not reflexive, it can't possible be transitive.

Edit: as pointed out in the comments, it is possible to have relations for which there is never any $ v \sim u$, even if $u \sim v$, such as with the relation $<$. In this case though, that's obvious so we're good.

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    $\begingroup$ This is almost true, but you also need some $u \sim v$ (plus either symmetry or $v \sim u$). If the relation is empty, then it is vacuously transitive and irreflexive. You can also have an asymmetric transitive irreflexive relation such as $<$ on the reals. $\endgroup$
    – Kevin
    Sep 3 '17 at 4:16
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For a pair of vectors, linear dependence means that one is a scalar multiple of another. With that in mind, take $w=\lambda u$, $\lambda\ne 0$. It should be pretty obvious that any vector $v$ that’s not a scalar multiple of $u$ isn’t a scalar multiple of $w$, either.

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