Let $V$ be a f.d inner product space with subspace $M,N$ and corresponding orthogonal projections $Q,P$. I need to prove that if $\| Px-Qx \| <\|x \|$ for all nonzero $x$, then $\dim M=\dim N$. As a hint, I was told to prove $P|_M$ and $Q|_N$ are injective.
I am totally lost. First of all, I don't have geometric intuition for this. Second, I don't see how the hint would help, and finally, I don't see how to prove it...
Help!
For geometric intuition, let $S$ be the sphere whose diameter is the segment $(0,x)$. Since the segments $(0,Px)$ and $(Px,x)$ are orthogonal, $Px$ lies on $S$, and similarly for $Qx$. Thus $\|Px-Qx\|\leq\|x\|$, with equality iff the segment $(Px,Qx)$ is also a diameter of $S$. This happens iff the four points are coplanar and form a rectangle, which implies $x=Px+Qx$, so $Px\in\ker Q|_N$ and $Qx\in\ker P|_M$.
Algebraically, if $x\in\ker P|_M$ is nonzero, then $x\in M$, so $Qx=x$ and your given inequality becomes $$ \|x\|>\|Px-Qx\|=\|x\|, $$ a contradiction. Thus $\ker P|_M=0$. That is, $P|_M$ is an injection from $M$ to $N$, so $\dim M\leq\dim N$. The reverse inequality follows similarly.