Given a complex square matrix $\alpha$, if $\alpha \alpha^\dagger$ is a real symmetric matrix, can one show that $\alpha$ can be written as $$ \alpha = \alpha_1 U $$ where $\alpha_1$ is real and U is a diagonal unitary matrix (only containing phase factors in the diagonal)?
The authors of https://arxiv.org/abs/0905.2562 (right after eq (56)) claim that this is possible, without an explanation.
I see that if the statement $ \alpha = \alpha_1 U $ is true, then each column of $\alpha$ is a real vector with an overall complex phase and hence each term in the matrix multiplication in $\alpha \alpha^\dagger$ is independently real. So the converse is trivially true. This approach however didn't lead me to a proof of the original statement.
All help is appreciated.
PS: It is possible that this statement is not true in general and the authors have used some more properties of these specific matrices which pertain to the physical problem. So if that's the case in your opinion, please write that too.
This statement is not in general true. Consider the following matrix: $$\alpha = \begin{bmatrix}6/5 & 8i/5 \\ -16i/5 & -12/5\end{bmatrix}, \text{ } \alpha \alpha^\dagger = \begin{bmatrix}4 & 0 \\ 0 & 16\end{bmatrix}$$ I created the counterexample as follows. Consider the spectral decomposition of $\alpha \alpha^\dagger$. Clearly if $\alpha \alpha^\dagger = U D U^\dagger$ for real $U$, then it will be symmetric and real, as desired. Recalling the singular value decomposition, we can see that if $\alpha = U D^{1/2} V^\dagger$ for unitary $V$ then $\alpha \alpha^\dagger = U D U^\dagger$ ($D^{1/2}$ is well-defined since $\alpha \alpha^\dagger$ is positive semidefinite for any $\alpha$). I chose $$U = \begin{bmatrix}1 & 0 \\ 0 & -1\end{bmatrix}, \text{ } D = \text{diag}(2, 4), \text{ } V^\dagger = \begin{bmatrix}3/5 & 4i/5 \\ 4i/5 & 3/5\end{bmatrix}$$ which contradicts the general proposition you hypothesized. It's clear that in general, if you don't pick $V^\dagger$ to be the product of a real unitary (orthogonal) matrix and a unitary diagonal matrix, then you have a counterexample of this form.