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Hey guys, I need to prove or refute that once given an eigenvalue t of a matrix AB and B is invertible,

so t is also eigenvalue of A.

I believe it's not true, but sadly beliefs are not enough in math : )

Thank you.

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    If you believe it's not true, you should try to come up with a counterexample. Here's a hint: try the simplest thing that could possibly work—let $A = [a]$ and $B = [b]$ be $1\times 1$ matrices.2011-03-22

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Let $A=B^{-1}$, then $t=1$. Clearly there are invertible matrices $B$ which don't have $1$ as an eigenvalue.

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    If A=B^-1, Do A and B have the same eigenvalues?2011-03-22
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    No, if $\{ \lambda_1, \dots , \lambda_n \}$ is the set of eigenvalues of $B$, then $\{ 1/\lambda_1, \dots , 1/\lambda_n \}$ is the set of eigenvalues of $B^{-1}$.2011-03-22
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    I see, and from this we get that 1 cannot be a eigenvalue of A also. Thanks.2011-03-22