Multiplicity of eigenvalues

1. It is known that the characteristic polynomial of $A\overline A$ is equal to the characteristic polynomial of $\overline AA$. Then $$ {\det(\overline AA-\lambda I)}=\det(A\overline A-\lambda I) $$ but for real numbers $\lambda$ the conjugate of the LHS is equal to the RHS, therefore the characteristic polynomial of $A\overline A$ has real coefficients.

2. An idea would be to prove that the characteristic polynomial of $A\overline A$ takes positive values for $\lambda \geq 0$. Then, if there exists a negative eigenvalue $\mu&lt0$ with odd multiplicity, (we can consider $\mu$ to be the the greatest such eigenvalue) then for $\lambda\in (-\mu-\varepsilon,-\mu) \cap [0,\infty)$ (for some $\varepsilon$ small enough) the polynomial would take negative values and we have a contradiction. It remains to prove that $$ \det(A\overline A-\lambda I)\geq 0,\ \forall \lambda \geq 0 $$ One way to do this is to prove that all the coefficients of the characteristic polynomial are positive.

here's a try:

(edit, deleted my development for q1 which was wrong, my bad!, thanks to Henning Makholm)

• If I'm not mistaken, you can show that the determinant is positive (because I think that you can show something like $\det(A\overline{A})=\det(A)det(\overline A)=\overline{\det A}^2$) which thus implies that the product of eigenvalue be positive. The product of $\lambda$ and $\overline{\lambda}$ is obviously positive if $\rm{Im}\lambda\neq 0$ but if you have a $\lambda\in\mathbb R$ and $\lambda&lt0$ then there must be another $\lambda'\in\mathbb R$ with $\lambda'&lt0$. This implies that there be an even number of real negative eigenvalues.

(edit, as Beni mentioned, this just shows that the total number of real negative eigenvalues is even (product is positive) which might not answer the question)