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The matrices $AB$ and $BA$ are similar provided either $A$ or $B$ is invertible.
If both $A$ and $B$ are singular (and square), a limiting argument involving $A +\epsilon I$
is useful. In this case $AB$ and $BA$ still have the same eigenvalues with the same
multiplicities. However, in general, $AB$ is not similar to $BA$, . Their Jordan forms may be different, in the sizes of the blocks associated with the eigenvalue $0$. |
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