# Simultaneous diagonalization of hermitian and anti-hermitian matrices

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.

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.