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 $\overline{\lambda}$.

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


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