Suppose two square $n\times n$ matrices, $A$ and $B$, do not commute. $$[A,B] = 0$$ However, suppose their commutator has a block of all 0's, i.e. \begin{align} [A, B] = \begin{pmatrix} * & \dotsm & * & * & \dotsm&*\\ \vdots & \ddots & \vdots & \vdots& \ddots&\vdots \\ * & \dotsm & * & *&\dotsm & *\\ * & \dotsm & * & 0&\dotsm & 0\\ \vdots & \ddots& \vdots & \vdots &\ddots & \vdots\\ *& \dotsm & *& 0&\dotsm & 0\\ \end{pmatrix} \end{align}
then how can this be interpreted in terms of the respective eigenvectors of the matrices? In trying to understand this I reviewed the accepted answer to "Why does a diagonalization of a matrix B with the basis of a commuting matrix A give a block diagonal matrix?". Here's my thinking so far,
Suppose $v$ is an eigenvector of $B$ with eigenvalue $\lambda$ lying in the subspace $V_0 = (v_k, v_{k+1}, ..., v_n)$ that corresponds to the $\mathbf{0}$ (zero matrix) block of $[A,B]$. Then \begin{align} [A,B]v = (AB-BA)v \end{align} will necessarily be orthogonal to $v$, and be supported entirely in the subspace $V_1 = (v_1, v_{2}, ..., v_{k-1})$.
However we cannot make the analogous statement to the fully commuting case, like that the vector $v'=Av$ is an eigenvector of $B$ with the same eigenvalue as $\lambda$. Maybe this isn't the right meaning of "commuting block"?
