Two finite-rank projections P and Q, if $\|P - Q\| < 1$, then rank $P$ = rank $Q$

Question

Let $P, Q \in \mathcal{L}(X)$ be two finite-rank projections. If $\|P - Q\| < 1$, then rank $P$ = rank $Q$.

My attempt:

If $P$ has rank of $n$, then $Px=\sum_1^n \hat x_j (\ell, x)$, where $\hat x_j$ is a set of orthonormal basis.

I want to prove by contradiction. Suppose rank $P$ > rank $Q$, then somehow $\|P-Q\| >= 1$.

I fail to use the property $P,Q$ are projections ($P^2x = Px$). Any help is appreciated.

Answer 1

HINT:

Since $||P-Q||<1$, the operator $I + (P-Q)$ is invertible. Therefore, for every $y \in X$ there exists $x \in X$ so that $x + (P-Q) x = y$, or $(x-Qx) + Px = y$.

It follows that $\ker Q + \operatorname{im} P = X$ and so rank $Q \le$ rank $P$.

$\bf{Added:}$

From the equation above we have $Qy = Q Px$. Therefore, the map $Q\colon \operatorname{im} P \to \operatorname{im} Q$ is surjective. So rank $Q \le$ rank $P$.

For the invertibility of $I + (P-Q)$, check Neumann series.

Answer 2

If say rank$(P) > $rank$(Q)$, you can find a nonzero element $x$ of the range of $P$ which is orthogonal to the range of $Q$. For in terms of basis elements of $Q$ the statement that $<x,y>$ $= 0$ for all $y$ in $Q$ reduces to an equation of rank$(Q)$ variables. If rank$(P) >$ rank$(Q)$ then there will be a nonzero solution.

In this situation, $(P - Q)x = Px - Qx = Px - 0 = x$. Thus $||P - Q|| \geq 1$.