Is linear independence of vectors transitive? 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?
 A: No.  In the plane, let $\mathbf{u} = \langle 1, 0 \rangle$, $\mathbf{v} = \langle 0, 1 \rangle$, and $\mathbf{w} = \langle 0, 2 \rangle$.
A: 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. 
A: 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.
A: 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.
