How do I show that if $\Vert Px-Qx \Vert <\Vert x \Vert$ for any $x\in V$ not $0$, then $\dim\left(M\right)=\dim\left(N\right)$.
$V$ is an inner product space and $M, N$ are sub-spaces of $V$.$P$ is a projection on $N$ and $Q$ is a projection on $M$.
I was given a clue, which says that I need to show that the projections $P|_N$ and $Q|_M$ are injective. (or maybe it is $P|_M$ and $Q|_N$, I'm not sure).
Anyway I don't understand how to approach to this question.
Hint: Indeed, $P|_M$ must be injective. Otherwise, there is an $x \in M$ for which $Px = 0$, which would imply that $\|Px - Qx\| = \|0 - x\| = \|x\|$.