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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 $B$.

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    $\begingroup$ What happens if $v$ is orthogonal to $x$? $\endgroup$ Jul 14 '20 at 11:45
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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 $\|x\|^2$.

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