Say I have two diagonalisable matrices $A$ and $B$, diagonalisable in different basis'. \begin{equation} A=\sum_ia_i|a_i\rangle\langle a_i|\qquad\text{and}\qquad B=\sum_ib_i|b_i\rangle\langle b_i|\;. \end{equation} Clearly both are diagonalisable. Therefore $A\otimes B$ and $B\otimes A$ are both diagonalisable. I know that in general the sum of two diagonalisable matrices is not diagonalisable, however I was wondering if because of the symmetry involved in this problem, the matrix: \begin{equation} A\otimes B+B\otimes A \end{equation}
might be diagonalisable?
I think this might help. First note that the square root of a diagonal matrix is diagonal and the inverse of a diagonal matrix is diagonal. The product of two diagonal matrices is diagonal. Assuming nothing about $A\otimes B+B\otimes A$ we can multiply by $\sqrt{A}^{-1}\otimes\sqrt{A}^{-1}$ on both sides leaving \begin{equation} \mathcal{I}\otimes\sqrt{A}^{-1}B\sqrt{A}^{-1}+\sqrt{A}^{-1}B\sqrt{A}^{-1}\otimes\mathcal{I}\;. \end{equation} $\sqrt{A}^{-1}B\sqrt{A}^{-1}$ must be diagonal as $\sqrt{A}^{-1}$ and $B$ are both diagonal. We can therefore say $C=\sqrt{A}^{-1}B\sqrt{A}^{-1}$, where $C$ is a diagonal matrix. Then $\mathcal{I}\otimes C\otimes\mathcal{I}$ must be diagonal. If $A\otimes B+B\otimes A$ is diagonal this must be true, however this can also be true if $A\otimes B+B\otimes A$ is not diagonal.