Convert Hermitian matrix to symmetric

Question

Is there some simple transformation (or a simple way to find it) which would convert any given Hermitian matrix $A$ to a symmetric matrix $B$ with the same spectrum as that of $A$ (so I guess that transformation should be unitary)?
I know I can represent an hermitian matrix as a $2\times2$ block matrix like $$A\to\left(\array{\Re A&-\Im A\\ \Im A&\Re A}\right),$$ but I'd like to preserve dimensions of $A$.

By simple I mean that I shouldn't have to diagonalize $A$ to achieve what I need (diagonalization is in fact what I want to do with the result, not having to take 4 times more space in complex case).

If not in general, then is there maybe some special-case way for such $A$ that $\Im A$ is a diagonal matrix?

Answer 1

Perhaps, the following might help: you can express $A$ as: $$A=B+iC$$ where $B$ is a symmetric and real matrix and $C$ is an anti-symmetric and real matrix, you can also guess for the eigenvectors in the same way: $$x=u +iv$$ Therefore, the eigenproblem becomes: $$(B+iC)(u+iv)=\lambda (u+iv)$$ That considering real and imaginary parts can be expressed as in your expression above: $$\begin{pmatrix}B & -C \\ C & B \end{pmatrix}=\lambda \begin{pmatrix} u \\ v\end{pmatrix}$$ But as you can notice, the dimension of the matrix is twice the dimension of the original problem and this fact is unavoidable.