prove two commutative Hermitian matrices have the same eigenvectors

xuphys

Hi,

Does anyone know how to prove that two commutative Hermitian matrices can always have the same set of eigenvectors?

i.e.

AB - BA=0

A and B are both Hermitian matrices, how to prove A and B have the same set of eigenvectors?

Thanks!

Fredrik

Since this is a textbook-style problem, the forum rules tell us to treat it as homework (even if it's not). So we can only give you hints, and you will have to show us what you've got so far. At the very least, you're going to have to show us that you know what the statement "v is an eigenvector of A" means.

xuphys

Thanks for your reply.

What I would say is that this is a common mathematical theorem which is one of the mathematical basis in quantum mechanics, but not a textbook style problem. I ask here because my text book (which is not written in English) only gives a simplified version of proof (assuming there is no duplicated eigenvalue). I am curious about a relatively more robust way to prove it.

Fredrik

Textbook problems often ask the reader to prove a theorem that wasn't proved in the text. So this certainly could be a textbook problem or a small part of a homework assignment.

micromass

I have moved this to homework. xuphys, please make an attempt when asking a question. What do you think of the problem? Is there something you can do?

xuphys

OK. What I thought is, since A and B are commutative, and they are also hermitians, so AB is also a hermitian (easy to prove).

Then AB is diagonalizable:

i.e.

Now I am unable to prove that both U^-1AU and U^-1BU are diagonal... and I am not sure whether I am on the right track. Could you help me with this?

Thanks!

fortissimo

Hermitian matrices are diagonalizable, so we have n eigenpairs (a_i, v_i) for A and eigenpairs (b_i, u_i) for B (where v_i and u_i may be chosen linearly independent).
It follows that A*v_i = a_i*v_i. This implies B*A*v_i = a_i*B*v_i. But this equals A*B*v_i, so B*v_i is an eigenvector to A with eigenvalue a_i. So B*v_i = k_i*v_i if eigenvalues are unique.
The argument is symmetric, so it follows that A and B have the same eigenvectors. The argument is more difficult if eigenvalues aren't unique, but you get invariant subspaces and block matrices, for which you can choose diagonal bases, roughly.