Projection with same rank

Question

Let $H$ be a Hilbert space (over $\mathbb{R}$ or $\mathbb{C}$), and $P$, $Q$ are two projections over $H$ such that $\|P-Q\|<1$. I am searching for a simple proof for $$ \dim\operatorname{Ran}P=\dim\operatorname{Ran}Q,\quad \dim\operatorname{Ran}(I-P)=\dim\operatorname{Ran}(I-Q). $$

To handle with this, we can define $R=(P-Q)^2$ and $U_1=QP+(I-Q)(I-P)$, $V_1=PQ+(I-P)(I-Q)$. Then it is easy to verify that $$ U_1V_1=V_1U_1=I-R,\quad U_1^*=V_1,\quad V_1^*=U_1. $$ Since $\|R\|<1$, $T=(I-R)^{-\frac12}$ makes sense (e.g. define it with series) and it commutes with $P$, $Q$. Set $$ U=U_1T,\quad V=V_1T, $$ then $UV=VU=I$ and $$ Q=UPU^{-1},\quad P=VQV^{-1}. $$ but this proof is too long and complicated, compared with the given condition. Therefore, I want to find an intuitive way to show that.

Answer 1

The assumption implies that $P$ is injective on ${\rm Ran}\,Q.$ Indeed assume that $Pv=0,$ $Qv=v\neq 0.$ Then $$\|v\|=\|Pv-Qv\|<\|v\|$$ a contradiction. As $P({\rm Ran}\,Q)\subset {\rm Ran}\,P$ we get $\dim {\rm Ran}\,P\ge \dim {\rm Ran}\,Q.$ By symmetry we get the converse inequality.

The same conclusion holds for $P'=I-P$ and $Q'=I-Q,$ as $\|P'-Q'\|<1.$

Since the ranks are equal there is an invertible operator $U$ such that $U({\rm Ran}\,P)={\rm Ran}\,Q$ and $ U({\rm Ran}\,(I-P))=\rm Ran(I-Q).$ We get $P=U^{-1}QU.$ The operator $U$ is unitary if $P$ and $Q$ are orthogonal projections.

Answer 2

Here is a proof by contrapositive.

Suppose $\dim \operatorname{ran}(P) \neq \dim \operatorname{ran}(Q)$. Without loss of generality, suppose $\dim \operatorname{ran}(P) > \dim \operatorname{ran}(Q)$. As a consequence (as I prove below), we can deduce that $\dim(\operatorname{ran}(P) \cap \operatorname{ran}(Q)^\perp) > 0$, so that there exists a non-zero vector $x \in \operatorname{ran}(P) \cap \operatorname{ran}(Q)^\perp$. From there, it suffices to note that $\operatorname{ran}(Q)^\perp = \ker(Q)$, so that $$ \|(P - Q)x\| = \|Px - Qx\| = \|x - 0\| = \|x\|, $$ so that $\|P - Q\| \geq 1$.

Thus, $\dim \operatorname{ran}(P) \neq \dim \operatorname{ran}(Q)$ implies that $\|P - Q\| \geq 1$. Similarly, $\dim\operatorname{ran}(I - P) \neq \dim \operatorname{ran}(I - Q)$ implies that $\|(I - P) - (I - Q)\| = \|P - Q\| \geq 1$. The conclusion follows.

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Claim: For closed subspaces $U,V \subset H$, $\dim(U) > \dim(V)$ implies that $\dim(U \cap V^\perp) > 0$.

Proof: Consider the quotient space $Q = H/V^\perp$, noting that $\dim(Q) = \dim(V)$. Let $\pi: H \to Q$ denote the associated projection map. Now, consider the restriction $\phi = \pi|_U:U \to Q$. Note that $$ \dim \operatorname{ran}(\phi) \leq \dim Q = \dim V. $$ Note that $\ker\phi = U \cap V^\perp$. On the other hand, $\phi$ induces an injective map $\tilde \phi:U/\ker\phi \to Q$, with $\operatorname{ran}(\tilde\phi) = \operatorname{ran}\phi$. Thus, we have $$ \dim (U/\ker \phi) = \dim(\text{ran}(\phi)) \leq \dim V \implies\\ \dim(U) \leq \dim(V) + \dim \ker \phi. $$ Because $\dim(V) < \dim(U)$, we are forced to conclude that $\dim \ker \phi = \dim (U \cap V^\perp)$ is positive, which is what we wanted.