Let $X$ and $Y$ be finite dimensional normal operators. There is a well-known result that, if $[X,Y] = 0$, then
- $X$ fixes the eigenspaces of $Y$, and $Y$ fixes the eigenspaces of $X$.
- There exists an orthonormal basis of eigenvectors that diagonalize both $X$ and $Y$ simultaneously.
Thus, if I want to know if $X$ and $Y$ share an eigenbasis, I can just compute $[X,Y]$, instead of actually computing the eigenvalues and eigenvectors.
I would like to know if one can find an algebraic condition on $X$ and $Y$ that would tell me if $X$ and $Y$ satisfy the much stronger condition that they share eigenspaces, rather than just an eigenbasis. This would mean that, in any eigenbasis of $X$ (resp, $Y$), we would have that $Y$ (resp $X$) is diagonal.