Operator theory: nonnegative operator $Q$ less than an orthogonal projection $P$ satisfies $PQ=QP$

Question

Suppose that we have a Hilbert space $\mathcal{H}$, an orthogonal projection $P$ (i.e. $P=|\Psi\rangle\langle\Psi|$ for some $\Psi\in\mathcal{H}$ with $\|\Psi\|=1$) and another non-negative bounded linear operator (not necessarily a projection!) $R$ satisfying $R\leq P$.

In my lecture notes, a lemma states that $R=RP=PR$, which is used to prove a theorem about joint measurability of POVMs in quantum mechanics. The proof is as follows:

1. Let $I$ denote the identity operator on $\mathcal{H}$, and $R\leq P$ implies $(I-P)R(I-P)=0$;

2. Because non-negative bounded operators have square roots due to the spectral theorem, we have $\sqrt{R}(I-P)=0$. Applying $\sqrt{R}$ to the left gives $R(I-P)=0$, hence $R=RP$ and similarly $R=PR$.

What I do not understand: why does $R\leq P$ imply $(I-P)R(I-P)=0$ ?

I consulted the following questions, but I am still stuck:

• Norm of orthogonal projection operator $P$, if $\text{Im}\,P\subseteq\text{Im}\,Q$, with $Q$ also an orthogonal projection
• Orthogonal Projections in Hilbert space

Answer 1

This result relies on the following general fact:

Lemma. Given $A\in\mathcal B(\mathcal H)$, if $A\geq 0$ and $A\leq 0$, then $A=0$.

This it suffices to show that the assumptions on $R,P$ imply that $(I-P)R(I-P)$ is at the same time positive and negative semi-definite.

• Because $R\geq 0$ and $P$ is self-adjoint (thus $I-P$ is also self-adjoint), $$ \langle x,(I-P)R(I-P)x\rangle=\langle (I-P)x,R(I-P)x\rangle\geq 0\,. $$
• Because $R\leq P$ (i.e. $P-R\geq 0$), using that $X\leq Y$ implies $Z^*XZ\leq Z^*YZ$ for all $X,Y,Z\in\mathcal B(\mathcal H)$ we compute $$ (I-P)R(I-P)\leq (I-P)P(I-P)=(P-P^2)(I-P)=0\,. $$