$P$, $Q$ are projections on a Hilbert space such that $|P-Q|<1$ then $\dim(\operatorname{Range}(P))=\dim(\operatorname{Range}(Q))$

Question

$P,Q$ are projections on a Hilbert space such that $|P-Q|<1$ then $\dim(\operatorname{Range}(P))=\dim(\operatorname{Range}(Q))$.

Is there a intuition for this to happen, also I am not being able to solve it.

Answer 1

I suppose you are considering finite rank projections. Let their ranges be $M$ and $N$. Define $T: M \to N$ by $Tx=Qx$. Then $T$ is linear. If $Tx=0$ with $\|x\|=1$ then $\|Px-Qx\|=\|x-0\| =\|x\|=1$ and $\|Px-Qx\| \leq \|P-Q\|<1$ which is a contradiction . Hence $T$ is a one-to-one linear map from $M$ into $N$. Since $M$ and $N$ finite dimensional this implies $dim(Range(P)) \leq dim(Range(Q))$. Now just reverse the roles of $P$ and $Q$ to get the reverse inequality.