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Suppose $A$ and $B$ are two $n \times n$ real matrices, with the special property that both $A$ and $B$ only have one eigenvalue, say equal to a real number $\lambda \ne 0$. That is, if $Av = \mu v$ for any complex number $\mu$, then actually $\mu = \lambda$, and similarly for $B$: if $Bw = \sigma w$ for some complex number $\sigma$ then actually $\sigma = \lambda$.

What can you say about the eigenvalues of $A^i B^j$ for $i,j \in {1,2,\dots}$? In general, the answer is nothing I think, what extra conditions might one need to put on $A$ and $B$ in order to say something?

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@Arturo: Wasn't the OP talking about $n\times n$ matrices? – Fabian Apr 10 '11 at 19:12
@Fabian: He edited the post; when I wrote my comment, he said $2\times 2$. The edit doesn't show because it was done within the first five minutes. – Arturo Magidin Apr 10 '11 at 19:20
Same eigenvalue for both? – Arturo Magidin Apr 10 '11 at 19:20
up vote 2 down vote accepted

In general $\det(A^i B^j) = \lambda^{(i+j)n}$, so that tells you the product of the eigenvalues, counted by multiplicity.

If $T(i,j) = \hbox{Trace}(A^i B^j)$, which is the sum of the eigenvalues of $A^i B^j$, counted by multiplicity, then $T(i,j)$ satisfies recurrence relations $\sum_{k=0}^n {n \choose k} (-\lambda)^{n-k} T(i,j+k) = 0$ and $\sum_{k=0}^n {n \choose k} (-\lambda)^{n-k} T(i+k,j) = 0$.

In particular, in the case $n=2$, the eigenvalues are determined by the determinant and the trace, and $T(i,j)$ is determined by the recurrence relation and $T(1,1)$: if $T(1,1) = \beta$, then I get $T(i,j) = ij{\lambda}^{i+j-2}\beta- \left( 2\,ij-2 \right) {\lambda}^{i+j}$

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Not a complete answer, just some observations:

If either $A$ or $B$ are diagonalizable, then they are a scalar multiple of the identity; if, say, $A=\lambda I$, then $A^iB^j = \lambda^iB^j$, which has eigenvalue $\lambda^{i+j}$. Symmetrically if $B$ is diagonalizable.

If the eigenvalue of $A$ is $0$, then $A^n = 0$, so $A^iB^j$ is the zero matrix for all $i\gt n$; similarly if the eigenvalue of $B$ is $0$.

If $A$ and $B$ commute and $\lambda\neq 0$, then $A$ maps eigenvectors of $B$ to eigenvectors of $B$ and symmetrically $B$ maps eigenvectors of $A$ to eigenvectors of $A$. As Robert Israel notes in comments, then $\lambda^{i+j}$ is an eigenvalue of $A^iB^j$, and is the only eigenvalue of $A^iB^j$.

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That's not quite right. If A and B commute, the eigenvalue of $A^i B^j$ is $\lambda^{i+j}$, not $\lambda^i$. And this is the only eigenvalue of $A^i B^j$: note that $A = N + \lambda I$ and $B = M + \lambda I$ where $N$ and $M$ are nilpotent and commute. Now $A B = N M + \lambda N + \lambda M + \lambda^2$. It's easy to see that $N M + \lambda N + \lambda M$ is nilpotent. Use induction to treat $A^i B^j$. – Robert Israel Apr 10 '11 at 20:05
@Robert: You're right. Thanks. – Arturo Magidin Apr 10 '11 at 21:16

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