Question on projections in linear algebra

Question

Hello all I was given this question for linear algebra class which I am stuck on and would truly appreciate the help:

$V$ is a finite-dimensional inner product space with $M$ and $N$ non trivial subspaces. $P$ and $Q$ are projections on $M$ and $N$ respectively and also for all non zero vectors $ 0 \neq x \in V $ we have: $ ||Px-Qx|| < ||x|| $. Show $\dim(M)=\dim(N)$.

I am given the hint that I should show the restriction of the projections on the subspaces is one to one.

I tried to proceed but cannot even deduce where to start from that is how to use the inequality given and projections be projections so I am truly stuck and would appreciate the help Thank you all.

Answer 1

Take $x\in M$, $x\ne 0$. Then $\|x-Qx\|<\|x\|$. Since $\|Qx\|\le \|x\|$, this shows that $Qx\ne 0$. Hence $Q|_M$ is injective and $\dim im(Q|_M)=\dim M$. Since $Q|_M$ is a restriction of $Q$, it follows $\dim im(Q)\ge \dim im(Q|_M)$. Now $M$ was the image of $P$, and hence $\dim im(Q)\ge \dim im (P)$.

Analogously you obtain the reverse direction.

This shows $\dim N=\dim im(Q)= \dim im (P)=\dim N$.