Say Matrix $A$ has eigenvalues $a_1,…,a_n$ and $B$ has eigenvalues $b_1,…,b_n$ i.e. $A$ and $B$ are diagonalizable.
Can there be made a statement about the eigenvalues of $AB$ ?
By the determinant we have ${\rm det}(A)=a_1 \cdots a_n$ and ${\rm det}(B)=b_1 \cdots b_n$, so ${\rm det}(AB)=a_1 \cdots a_n\cdot b_1 \cdots b_n$, but that does not imply that the eigenvalues are $a_1 b_1, … , a_n b_n$.
In fact if $U$ diagonalizes $A$ with diagonal matrix $\Lambda_a$ and $V$ diagonalizes $B$ with diagonal matrix $\Lambda_b$ then
$$
\Lambda_a \Lambda_b = U^{-1} A U \, V^{-1} B V
$$
or vice versa
$$
A B = U \Lambda_a U^{-1} \, V \Lambda_b V^{-1}
$$
but this does not give me any information in general about the eigenvalues of $AB$.
If $A$ and $B$ commute, then they can be diagonalized simultaneously with $U=V$ and then
$$
AB=U \Lambda_a \Lambda_b U^{-1}\, .
$$
So is there some other way to simply deduce the eigenvalues of $AB$ ?
EDIT: So as far as the answers go, there does not in general seem to be a method to deduce the eigenvalues of $AB$ simply from $A$ and $B$.
Can there be made statements about the maximal eigenvalues of $AB$. One was already made i.e. $$\rho(AB)\leq\rho(A)\rho(B)$$
where $\rho$ is the spectral radius, but that is only an inequality. Is there an equality for the maximal eigenvalue?
Best Answer
Not much can be said. For instance, suppose we define $M_{\theta}$ as
$ \left[ \begin{matrix} 1+cos(\theta) &sin(\theta) \\ sin(\theta)& 1-cos(\theta)\\ \end{matrix}\right]$
The eigenvalues of $M_{\theta}$ will be $0$ and $2$ regardless of the value of $\theta$.
If we take $A =M_0$, $B=M_{\theta}$, then $AB$ is
$2 \left[ \begin{matrix} 1+cos(\theta) &sin(\theta) \\ 0&0\\ \end{matrix}\right]$
which has eigenvalues $0$, $2(1+cos(\theta))$. Thus, thus despite $A$ and $B$ having fixed spectra, the largest eigenvalue of $AB$ can range anywhere from $0$ to $4$.