Let $A_\alpha$ be a family of commuting matrices, that is, $A_\alpha A_\beta=A_\beta A_\alpha$. Show that there exists an unitary matrix $U$ such that $U^*A_\alpha U$ is upper triangular for each $\alpha$.
I know that there should be something related to invariant subspaces. However, I could not proceed…
Best Answer
Let ${\cal F}$ be the family of commuting matrices. Call a subspace $S$ invariant if $AS \subset S$ for all $A \in {\cal F}$.
Let $P_n$ be the proposition that the result is true for $n \times n$ matrices.
Here is the outline of one approach:
(1) Note $P_1$ is trivially true.
(2) Show that if $n>1$ and there exists a one dimensional invariant subspace then the problem can be reduced to $P_{n-1}$.
(3) Show that if $S$ is an invariant subspace with $\dim S >1$ then there exists an invariant subspace of strictly smaller dimension.
(4) Show that if $A \in {\cal F}$ and $\lambda$ is an eigenvalue of $A$, then $\ker (A-\lambda I)$ is invariant.