# Show $( v^\intercal x ) ^{2} = x^\intercal x$ given $\| v \| =1$

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 $B$.

• What happens if $v$ is orthogonal to $x$? Jul 14 '20 at 11:45

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 $$\|x\|^2$$.