1. Motivation
Consider symmetric matrices $A,B\in\mathbb{R}^{n\times n}$, and let $A$ be full-rank and $B$ be low-rank. The simultaneous block-diagonalization, defined as the following
$$A=V_{1}\hat{A}_{1}V_{1}^{T}+V_{2}\hat{A}_{2}V_{2}^{T},\qquad B=V_{1}\hat{B}_{1}V_{1}^{T},$$
is extremely useful because it simplifies the parabolic equation $\frac{d}{dt}x=Ax+Bu$ into two distinct and decoupled systems,
$$\frac{d}{dt}\left(V_{1}^{T}x\right) =\hat{A}_{1}\left(V_{1}^{T}x\right)+\hat{B}_{1}\left(V_{1}^{T}u\right),\\ \frac{d}{dt}\left(V_{2}^{T}x\right) =\hat{A}_{2}\left(V_{2}^{T}x\right).$$
Assuming commutivity, $AB=BA$, choosing $V_{1},V_{2}$ is a simple matter of partitioning the shared eigenspace of $A,B$. However, when A,B are non-commutative, their eigenspaces may be very different, and a block-diagonalization cannot be exactly achieved. We desire an approximation, in which the "strength" of the off-diagonal blocks is bounded with respect to the dominant diagonal blocks.
$$\|V_{1}^{T}AV_{2}\|_{F}\le\alpha\|V_{1}^{T}AV_{1}\|_{F},\qquad\|V_{1}^{T}BV_{2}\|_{F}\le\alpha\|V_{1}^{T}BV_{1}\|_{F}.$$
Hence we are motivated by the parabolic PDE context to pose the following question.
In the non-commutative case, i.e. $AB\neq BA$, how does one chose $V_{1},V_{2}$ to approximate a simultaneous block-diagonalization without setting $V_{2}=0$?
2. View via representation theory.
The same question can be posed in a more general manner using the language of representation theory. (Our original problem is recovered by writing $A_2:=V_2 V_2^T A V_2 V_2^T$ and $A_1:=A-A_2$.)
Given symmetric, non commutative matrices, $A,B$, is there a decomposition, $A=A_{1}+A_{2},$ such that $A_{2}\neq0$ is (approximately) commutative with $A_{1}$ and $B$, and that the rank of $A_{1}$ is minimized?
In other words, can we find an $A_{2}$ such that $A_{1}A_{2}\approx A_{2}A_{1}$ and $BA_{2}\approx A_{2}B$, while allowing for $BA_{1}\neq A_{1}B$?
In the commutative matrix context, it may be easier to redefine "approximate" w.r.t. the norm of the commutator,
$$AB\approx BA\text{ iff }\|AB-BA\|_{F}\le\alpha\|AB\|_{F}.$$
There is strong reason to believe that low-rank has an important role in this context. Choosing $A_2$ by projecting $A$ onto the kernel / nullspace of $B$, we are guaranteed to satisfy $A_2 B = B A_2=0$. However, it is unlikely for this to simultaneously satisfy $A_{1}A_{2}\approx A_{2}A_{1}$. Likewise, we can guarantee that $A_{1}A_{2} = A_{2}A_{1} =0$ by partitioning the eigenspace of $A$. However, in this case, there is no guarantee that $BA_{2}\approx A_{2}B$.