Why does $(\vec{v}\cdot\vec{u})\vec{u}\neq(\vec{u}\cdot\vec{u})\vec{v}$? I'm trying to save time running Gram-Schmidt. Why doesn't this product equal $||\vec{u}||\vec{v}$? More specifically (and I know this is fundamental and that I should already know it), why doesn't the dot product commute between more than two vectors?
 A: The inner product does commute, but scalar vector multiplication doesn't work with it. $(v\cdot u)u$ is in the direction of (parallel to) $u$ and $(u\cdot u)v$ is in the direction of $v$, thus if $u\nparallel v$, the vectors point in entirely different directions (let alone being equal).
In fact, the equality holds if and only if there is some $\lambda\in\mathbb R$ such that $u = \lambda v$ (or if $v=0$). Then
$$(v\cdot u)u = (v\cdot \lambda v)\lambda v = \lambda^2 \|v\|^2 v\\
(u\cdot u)v = (\lambda v \cdot \lambda v) v = \lambda^2 \|v\|^2 v$$
If they are not parallel, the two vectors can't be equal because equal vectors must be parallel and parallelity is an equivalence relation:
$$(v\cdot u)u \parallel u \nparallel v \parallel (u\cdot u)v \\
\Rightarrow (v\cdot u)u \nparallel (u\cdot u) v\\
\Rightarrow (v\cdot u)u \ne (u\cdot u)v$$
A: If $u = (1,0,0)$ and $v = (1,1,1)$, then $v\cdot u = 1$ and $u\cdot u = 1$. So $(v\cdot u)u = u$, but $(u\cdot u)v = v \neq u$.
If either $u$ or $v$ is the zero vector, equality holds. So assume neither $u$ nor $v$ is the zero vector. If equality holds, then $v = Au$, where $A = (v\cdot u)/(u\cdot u) \in \Bbb R$. Conversely, if $v = tu$ for some $t\in \Bbb R$, then $v\cdot u = tu\cdot u$, thus $(v\cdot u)u = t(u\cdot u)u = (u\cdot u)v.$ 
