So, I was doing homework and I had to turn an arbitrary invertible $N \times N$ matrix $A$ into a Hermitian $2N \times 2N$ matrix $A'$. This is rather simple, namely pick $A' = \begin{pmatrix} O & A^* \\ A & O\\ \end{pmatrix}$. However, I got curious about the eigenvalues. Are the eigenvalues of $A'$ easy to calculate from $A$? I personally expect so. Since the eigenvalues of $A^*$ are the complex conjugates of those of $A$, I thought this could somehow but used, but I am not entirely sure.
So, would this be possible?
Observe that \begin{align*} \det(A' - \lambda I) &= \det \begin{pmatrix} - \lambda I & A^* \\ A & - \lambda I \end{pmatrix} \\ &= \det(\lambda^2 I - A^* A) \end{align*} That is, the eigenvalues of $A'$ are the square roots of the eigenvalues of $A^* A = A^2$, which are precisely the singular values of $A$.