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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?

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    @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](http://math.stackexchange.com/questions/46544/why-does-a-diagonalization-of-a-matrix-b-with-the-basis-of-a-commuting-matrix-a) and a somewhat more general version is [here](http://math.stackexchange.com/questions/109811/).2012-03-08

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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.

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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).