CONVEX AND CONCAVE FUNCTIONS OF SINGULAR VALUES OF MATRIX SUMS
This question comes from the proof of theorem $2$ linked above.
The author firstly defines $|A|=(AA^{*})^\frac{1}{2}.$
If $H $ and $K$ are Hermitian matrices, the notation $H\le K$ means $K-H$ is a positive semidefinite matrix.
Theorem $2$ is : Let A and B be square matrices, not necessarily Hermitian. Then unitary matrices $U$ and $V$ exist such that $$|A+B|\le U|A|U^{*}+V|B|V^{*}.$$
In one part of the proof, the author claims :$$\frac{1}{2}(A+A^{*})\le U|A|U^{*}$$ for some unitary $U$.
He also says :take $U$ to be such that a unitary similarity simultaneously brings $\frac{1}{2}(A + A^{*})$ and $U|A|U^* = U(AA ^*)^\frac{1}{2}U^*$ into diagonal form with diagonal elements in nonincreasing order.
I think this may be a question about simultaneous diagonalization of $\frac{1}{2}(A + A^{*})$ and $(AA ^*)^\frac{1}{2}.$
But I don't konw how to proof that $\frac{1}{2}(A + A^{*})$ and $ (AA ^*)^\frac{1}{2}$ can be simultaneously diagonalizable by a unitary matrix.
Thanks!
I give the proof of the author in the form of pictures. Between Equation $(6)$ and the theorem 2's proof is a proof of a lemma.
The article does not claim that $\frac{1}{2}(A+A^*)$ and $(AA^*)^{\frac{1}{2}}$ are simultaneously diagonalizable. What the article claims is that there exists a unitary $U$ such that $\frac{1}{2}(A+A^*)$ and $U(AA^*)^{\frac{1}{2}}U^*$ are simultaneously diagonalizable. The latter ist easy to see as $$\frac{1}{2}(A+A^*) = V_1D_1V_1^* \quad \text{and} \quad (AA^*)^{\frac{1}{2}} = V_2D_2V_2^*$$ for some unitary $V_1,V_2$ and diagonal $D_1,D_2$. Now choose $U := V_1V_2^*$.