Simultaneous diagonalization of hermitian and anti-hermitian matrices

Question

I have an unitary matrix U, that i can decompose on the hermitian and anti-hermitian parts.

$$U = A+B$$

How to prove that two matrices are diagonalizable simultaneously without knowledge, that unitary matrices are diagonalizable.

I tried to solve this problem like this:

1)$S^HAS = D_1$, where $D_1$ - diagonal matrix

2)Then i have a matrix: $V = S^HBS $

3)I know, that all symmetric matrices are diagonazible, but V is not symmetric, so i don`t know what to do.

Answer 1

Note that the Hermitian and anti-Hermitian parts of $U$ can be written as "polynomials" (more accurately, rational functions) of $U$. In particular, we have $$ A = \frac 12 (U + U^*) = \frac 12 (U + U^{-1}), \quad B = \frac 12 (U - U^*) = \frac 12 (U - U^{-1}). $$ It follows that $A$ and $B$ commute. In particular, we can expand the product of the $\frac{U \pm U^{-1}}{2}$ matrices to get $$ AB = BA = \frac 14 (U^2 - U^{-2}). $$ Because $A,B$ are diagonalizable with $AB = BA$, it follows that $A$ and $B$ are simultaneously diagonalizable.