Show $( v^\intercal x ) ^{2} = x^\intercal x$ given that $v$ is a unit vector.
I believe that $v$ is ought to cancel out by $v^\intercal v = 1$ but don't see how:
$( v^\intercal x ) ^{2} = ( v^\intercal x )( v^\intercal x ) = \dots ?$
My attempt was to try something like $( v^\intercal x )( v^\intercal x ) = ( x^\intercal v ) ( v^\intercal x ) = x^\intercal ( v v^\intercal ) x$, but I get the outer product $v v^\intercal$ instead of $v^\intercal v$ so that's a bummer.
Background for this problem is my last question: Matrix norm inequality $\| Bx\| \geq |\lambda| \| x \|$ for a real symmetric $B$.
That cannot be shown. By Cauchy-Schwarz inequality, $(v^Tx)^2\le(v^Tv)(x^Tx)=x^Tx$. Equality holds if and only if $x$ is parallel to $v$. When $v$ and $x$ are linearly independent, $(v^Tx)^2$ is strictly smaller than $\|x\|^2$.