The operator matrix on Hilbert space

Question

Let $H$ be a Hilbert space and $P$ be the projection operator, then $H= P(H)\oplus (1-P)(H)$. Hence, for each $T\in B(H)$, we have $$T=\left(\begin{array}{ccc} PTP & PT(1-P) \\ (1-P)TP & (1-P)T(1-P) \\ \end{array}\right)$$ I hope to know why we can decompose an operator $T$ into such a form of matrix?

Answer 1

It is just a fancy (and useful) way of writing the trivial equality $$ T=PTP+(I-P)TP+PT(I-P)+(I-P)T(I-P). $$ It is useful because both addition and multiplication behave like the corresponding operations for matrices, so you can actually treat $T$ as a $2\times2$ matrix (without forgetting that the entries are operators).