Does $A$ commuting with a block-diagonal matrix $B=B_1\oplus B_2$ implies it is block-diagonal?

Question

Let $B$ be a block-diagonal matrix, $B\equiv B_1\oplus B_2$, and suppose that $A$ is a matrix such that $[A,B]=0$.

Does this imply that $A$ has the same block-diagonal structure as $B$ (that is, it is block-diagonal with blocks placed at the same positions as those of $B$)?

I understand that, as discussed in Why does a diagonalization of a matrix B with the basis of a commuting matrix A give a block diagonal matrix?, if $A$ and $B$ commute, then they respect each others eigenspaces. This would mean that if, say, $V_1$ is an eigenspace of $B$, then $A V_1\subseteq V_1$. Therefore, if $B_1$ is diagonalizable, then I think $A$ should at least have the block structure $$ A = \begin{pmatrix} A_{11} & A_{12} \\ 0 & A_{22}\end{pmatrix}.$$ However, what about the cases in which $B_i$ are not necessarily diagonalizable? Do similar conclusions hold? If not, what are some counterexamples?

I'm interested in the statement when we are not assuming the underlying field to be algebraically complete. Another realted question is Commuting block diagonal matrices, but they also seem to be making further assumptions about the diagonalizability of the underlying matrices.

Answer 1

An easy counterexample is $$ B=\left(\begin{array}{cc|cc}p&0&0&0\\ 0&1&0&0\\ \hline0&0&1&0\\ 0&0&0&q\end{array}\right), \ A=\left(\begin{array}{cc|cc}0&0&0&0\\ 0&1&1&0\\ \hline0&1&1&0\\ 0&0&0&0\end{array}\right) $$ for any $p$ and $q$.

However, the statement is true when $B_1$ and $B_2$ do not share any eigenvalue over the algebraic closure of the underlying field. In this case, when $A$ is comformingly partitioned as $\pmatrix{X&Y\\ Z&W}$, the Sylvester equations $B_1Y=YB_2$ and $B_2Z=ZB_1$ have only trivial solutions $Y=0$ and $Z=0$.