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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?

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$\textbf{Proposition}:$ Let $a,b,c\geq 0$ and $n\geq 2$. Then there are $A,B\in M_n(\mathbb{R})$ s.t. $\rho(A)=a,\rho(B)=b,\rho(AB)=c$.

$\textbf{Proof}$. Case 1. $a=b=0$. Take $A=diag(0_{n-2},\begin{pmatrix}0&0\\1&0\end{pmatrix}),B=diag(0_{n-2},\begin{pmatrix}0&c\\0&0\end{pmatrix})$.

Case 2. $a\not= 0$ or $b\not= 0$, for example $a\not= 0$. Take $A=diag(0_{n-2},\begin{pmatrix}0&0\\0&a\end{pmatrix}),B=diag(0_{n-2},\begin{pmatrix}p&q\\r&c/a\end{pmatrix})$ and use the fact that for every $P\in\mathbb{R}[x]$ of degree $2$, there are $p,q,r\in\mathbb{R}$ s.t. the characteristic polynomial of $\begin{pmatrix}p&q\\r&c/a\end{pmatrix})$ is $P$.

$\textbf{Remark}$. If $A,B$ are simultaneously triangularizable (for example when $rank(AB-BA)\leq 1$), then there are orderings $(\lambda_i),(\mu_i)$ of $spectrum(A),spectrum(B)$ s.t. $spectrum(AB)=(\lambda_i\mu_i)$.

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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 $4$.

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    $\begingroup$ The first statement is not true. Take $A=\begin{bmatrix}1 & n \\ 0 & 1 \end{bmatrix}$, $B=\begin{bmatrix}0 & -1 \\ 1 & 0 \end{bmatrix}$. Then then eigenvalues of both $A,B$ are bounded by one, but it is not hard to see that for large $n$, $AB$ has an eigenvalue close to $n$. $\endgroup$ – copper.hat Jul 26 '18 at 16:38
  • $\begingroup$ How is that possible, because Gefands corollary en.wikipedia.org/wiki/Spectral_radius#Gelfand_corollaries states that indeed $\rho(AB) \leq \rho(A)\rho(B)$. Or must the eigenvalues be real? Here the eigenvalues of $B$ are $\pm i$. $\endgroup$ – Diger Jul 26 '18 at 18:23
  • $\begingroup$ Or does the Theorem in fact need "diagonalizability" ? $\endgroup$ – Diger Jul 26 '18 at 18:41
  • $\begingroup$ @Diger Read that Wikipedia entry again. It says "assuming that they all commute ". In your case, the inequality is guaranteed to hold only when $AB=BA$, but that's evident because $A$ and $B$ are then simultaneously triangulable over $\mathbb C$. $\endgroup$ – user1551 Jul 27 '18 at 4:26
  • $\begingroup$ Indeed, I overread that which makes this statement somewhat rediculous though. $\endgroup$ – Diger Jul 28 '18 at 17:02
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Consider $A = aI, B = bI$, with $ab = 1$. Then the eigenvalues of $A$ and $B$ can be made arbitrarily different, while those of $AB$ are all $1$s.

In general, the largest and smallest evalues of $AB$ are bounded by the products of the largest and smallest (respectively) evalues of $A$ and $B$, but that's a very weak claim.

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  • $\begingroup$ You seem to be saying that given fixed eigenvalues for AB, the eigenvalues of A and B can vary arbitrarily. But a more on-point claim would be that given fixed eigenvalues of A and B, the eigenvalues of AB can vary arbitrarily. $\endgroup$ – Acccumulation Jul 26 '18 at 14:39
  • $\begingroup$ I agree; on the other hand, the second claim isn't true -- there are the max and min bounds I mention in my answer. A better example would have been $A = B = diag(a, b)$; now the eigenvalues of $AB$ vary quadratically in those of $A$ and $B$; if instead you make $B = diag(b, a)$, the the eigenvalues of $AB$ are all $1$s, so knowing the evalues of $A$ and $B$ tells you very little about those of $AB$. $\endgroup$ – John Hughes Jul 26 '18 at 16:11
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In general, no, for your setup you pretty much described the case where it is possible to deduce the eigenvalues of $AB$.

However, you still may get some bounds on $\lambda_{\min}(AB)$ and $\lambda_{\max}(AB)$.

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