# Eigenvalues of matrix product of two hermitian matrices

Show that the eigenvalues of the matrix product of two Hermitian matrices are either real or appear as pairs of complex conjugates, i.e., if $\lambda$ is an eigenvalue, then so is its complex conjugate $\overline{\lambda}$.

• Have you tried anything yet? – Exodd Jun 18 '15 at 14:20
• For those curious, an example where the product has complex eigenvalues: $$\pmatrix{-1\\&1}\pmatrix{0&1\\1&0}$$ – Omnomnomnom Jun 18 '15 at 15:24

We have $$\begin{split}\overline{\det(AB-\lambda I)} &= \det\left[(AB-\lambda I)^*\right] = \det(B^*A^*-\overline\lambda I) \\&= \det(BA-\overline\lambda I) = \det(AB-\overline\lambda I),\end{split}$$ where the last equality follows from the Sylvester's determinant theorem.