Commutant of a projection in $B(H)$

Question

Suppose $P$ is a projection in $B(H)$. How to compute the commutant of $P$. I saw a conclusion: The commutant $P'$ is $PB(H)P$.

It is easy to check that $PB(H)P\subset P'$, but how to verify the convese direction?

Answer 1

Consider the decomposition $H=PH\oplus(1-P)H$. With respect to this decomposition, $P=\begin{pmatrix}1&0\\0&0\end{pmatrix}$. If $T=\begin{pmatrix}a&b\\c&d\end{pmatrix}\in B(H)$ commutes with $P$, then $$\begin{pmatrix}a&b\\0&0\end{pmatrix}=PT=TP=\begin{pmatrix}a&0\\c&0\end{pmatrix},$$ so that $b=0$ and $c=0$.

Conversely, if $T=PaP+(1-P)d(1-P)$ for some $a,d\in B(H)$, then $T$ commutes with $P$. Thus, the commutant of $P$ is $$PB(H)P+(1-P)B(H)(1-P).$$