Eigenvalues of Product of 2 hermitian operators Let $A$ and $B$ be two Hermitian operators. Let $C$ be another operator such that $C = AB$. What can we say about Eigenvalues of $C$? Will they be real/imaginary/complex? What I did was to search for examples. The following were examples (in matrix representation) I looked for:
$ A =
\begin{bmatrix}
1 & 0\\
0 & 1
\end{bmatrix}$ and $ B = \begin{bmatrix}
0 & 1\\
1 & 0
\end{bmatrix}$ to get a hermitian matrix and so real eigenvalues.
Next I tried:
$ A = \begin{bmatrix}
1 & 0\\
0 & -1
\end{bmatrix}$ and $ B = \begin{bmatrix}
0 & 1\\
1 & 0
\end{bmatrix}$ to get Anti-Hermitian matrix and so imaginary eigenvalues.
Is there a more concrete way of solving this? Can we have a general complex number as eigenvalues for the product of the Hermitian Matrices?
 A: In general, we can say that $C=AB$ will have real, imaginary and complex eigenvalues (complex of the form $z=a+ib$ where and $\{a,b\in \mathbb{R}\mid a,b \ne 0\}$ as shown in the comments by Mark and Qmechanic's answer). For example, if
$$A=\begin{bmatrix}
0 &1 \\ 
 1& 0
\end{bmatrix}\ \ \text{and}\ \  B=\begin{bmatrix}
1 & 0\\ 
 0& -1
\end{bmatrix}$$ where $$AB=\begin{bmatrix}
0 &-1 \\ 
 1& 0
\end{bmatrix}$$ will not have real, but imaginary eigenvalues.
However, one thing we can say is that if $A$ and $B$ commute then $C=AB$ will always have real eigenvalues, since the eigenvalues of all Hermitian operators are real.
So if $$C=AB$$ then $$C^\dagger =(AB)^\dagger =B^\dagger A^\dagger =BA$$ since $A$ and $B$ are Hermitian, and clearly $$C^\dagger =C$$ if $$[A,B]=AB-BA=0\rightarrow AB=BA$$ This means that $C^\dagger =C$ only if $A$ and $B$ commute in which case $C$ will have real eigenvalues.
A: TL;DR: Assuming that $A,B$ are self-adjoint, the product $AB$ does not need to be diagonalizable. And if $AB$ is diagonalizable, the eigenvalues need not be real or imaginary.
Example 1: $AB$ is not diagonalizable:
$$A~=~\begin{pmatrix} 0 & 1 \cr 1 & 0 \end{pmatrix} \quad\wedge\quad B~=~\begin{pmatrix} 1 & 0 \cr 0 & 0 \end{pmatrix}\quad\Rightarrow\quad   AB~=~\begin{pmatrix} 0 & 0 \cr 1 & 0 \end{pmatrix}.$$
Example 2: $AB$ has complex eigenvalues:
$$A~=~\begin{pmatrix} 0 & 1 \cr 1 & 0 \end{pmatrix}\quad\wedge\quad B~=~\begin{pmatrix} 0 & b \cr b^{\ast} & 0 \end{pmatrix}\quad\Rightarrow\quad AB~=~\begin{pmatrix} b^{\ast} & 0 \cr 0 & b  \end{pmatrix}. $$
