Q: Exercise 2.2.3 in Analysis Now I have the next question
Let $Y$ and $Z$ be complementary of $X$. Show that every operator $T$ in $B(X)$ gives rise to four operators $T_{11}\in B(Y)$, $T_{22}\in B(Z)$, $T_{12}\in B(Z,Y)$ and $T_{21}\in B(Y,Z)$ such that $T$ can regarded as an operator matrix
$$
T=\begin{pmatrix}
T_{11} & T_{12}\\
T_{21} & T_{22}
\end{pmatrix}
$$
I proved that sum and product of elements in $B(X)$ are compatible with
matrix sum and product of the corresponding matrices. I don't know how to attack this: Give necessary and sufficient conditions on the matrix for having $T(Y)\subset Y$ or for $T(Z)\subset Z$.
I have tried this exercise, it is something similar but I do not know how to combine them.

Define the operator $P: X \to X$ by $Px = y$, where $x = y + z$. Show that
$P\in B(X)$ and that P is idempotent. Show, conversely, that if $P$ is an idempotent operator in $B(X)$, then $Y = P(X)$ and $Z = \ker P$ are complementary subspaces. Show finally that an operator $T in B(X)$ commutes with $P$ iff $T(Y)\subset Y$ and $T(Z)\subset Z$.

Any ideas?

A: Almost exactly what Ciaran O Raghaillaigh said at the comment section: $T(Y)\subset Y$ if-f $T_{12}=0$.
We write a vector $y$ of $Y$ in the form $(y,0)^t$, so, $Ty$ is the formal multiplication of the matrix $[T_{i,j}]$ with the column vector $(y,0)^t$. $Ty=T_{11}y+T_{12}y$. If $T_{12}=0$, then $Ty=T_{11}y\in Y$, because by definition $T_{11}\in B(Y)$. Conversely, if $Ty\in Y$ for all $y\in Y$, then $T_{12}y=Ty-T_{11}y\in Y$ (because $Ty\in Y$ and $T_{11}y\in Y$). But also $T_{12}y\in Z$, so $T_{12}y\in Y\cap Z=0$, so $T_{12}=0$.
Likewise, $T(Z)\subset Z$ if-f $T_{21}=0$.
