My question is:

If $A$ and $B$ are two Hermitian matrices, and $AB$ is also a Hermitian matrix, then how do prove that both $A$ and $B$ are diagonalizable through the same unitary matrix (i.e the unitary matrix that diagonalizes $A$, diagonalizes $B$ as well).

It is obvious that in order for $AB$ to be Hermitian, $A$ and $B$ have to commute, i.e: $AB=BA$. Can anyone tell me how to prove that the same unitary matrix that diagonalizes $A$, diagonalizes $B$ as well?

  • 1
    $\begingroup$ It's not quite true as you stated. For a stupid example, if $A$ is $I$ and $B$ is any non-diagonal Hermitian matrix, then $I$ diagonalizes $A$, but won't diagonalize $B$. What is true is that some matrix will diagonalize both. In the generic case when $A$ and $B$ both have distinct eigenvalues, then I think it's true that if a matrix diagonalizes $A$ it must diagonalize $B$, but I'll have to think about it a bit more. $\endgroup$ – Jason DeVito Feb 29 '12 at 2:39
  • $\begingroup$ @Jason: Yes, if $A$ has all distinct eigenvalues, then a matrix that diagonalizes $A$ will also diagonalize $B$. A matrix that commutes with a diagonal matrix with distinct diagonal entries is diagonal. A slight generalization of this came up here and a somewhat more general version is here. $\endgroup$ – Jonas Meyer Mar 8 '12 at 18:31

The orthogonal projections on eigenspaces of $A$ and of $B$ can be written as polynomials in $A$ and $B$ respectively, so they commute with each other and with $A$ and $B$. The nonzero products of an orthogonal eigenspace projection for $A$ and an eigenspace projection for $B$ are orthogonal projections on subspaces of ${\mathbb C}^n$ where $A$ and $B$ both act as multiples of the identity matrix. Take an orthonormal basis whose members are all in those subspaces, and the matrices for $A$ and $B$ in that basis will both be diagonal.


It may be useful to realize that the question/assertion can be reformulated (noting @Jason DeVito's comment) as that there is an orthogonal basis of simultaneous eigenvectors for two commuting self-adjoint (=hermitian) operators $S,T$ on a finite-dimensional space. This is a standard consequence of the fact that $S$ preserves the eigenspaces of $T$ (although not necessarily subspaces of the eigenspaces).


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.