**problem:**

Two matrices \(\displaystyle A,B\in\mathbb{C}^{m\times m}\) are unitarily equivalent if \(\displaystyle A=QBQ^{*}\) for some unitary \(\displaystyle Q\in\mathbb{C}^{m\times m}\).

Is it true or false that \(\displaystyle A\) and \(\displaystyle B\) are unitarily equivalent if and only if they have the same singular values?

**attempt:**

__First direction:__If \(\displaystyle A=QBQ^{*}\), do \(\displaystyle A\) and \(\displaystyle B\) have the same singular values?

Let,

\(\displaystyle A=U_A \Sigma_A V^{*}_A\), and let,

\(\displaystyle B=U_B \Sigma_B V^{*}_B\).

Since \(\displaystyle A=QBQ^{*}\), we have that;

\(\displaystyle QBQ^{*}=U_A \Sigma_A V^{*}_A \Rightarrow B=(Q^{*}U_A)\Sigma_A(V^{*}_AQ)\).

This shows that \(\displaystyle B\) has the same singular values as \(\displaystyle A\).

__If \(\displaystyle A\) and \(\displaystyle B\) have the same singular values, are they unitarily equivalent?__

Second direction:

Second direction:

Let,

\(\displaystyle A=U_A \Sigma V^{*}_A\), and let,

\(\displaystyle B=U_B \Sigma V^{*}_B\).

Then,

\(\displaystyle U^{*}_A A V_A = U^{*}_B B V_B\) and so,

\(\displaystyle A=(U_A U^{*}_B)B(V_B V^{*}_A)\).

I would like to show that \(\displaystyle U_BU^{*}_A=V_BV^{*}_A\), but do not know how.

Interesting fact is that \(\displaystyle AA^{*}=(U_AU^{*}_B)BB^{*}(U_BU^{*}_A)\). Maybe I could use that in some way?

All suggestions are appreciated, thanks.