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If $\lambda$ and $\mu$ are eigenvalues of two $n \times n$ matrices $A$ and $B$ respectively, then $\mu\lambda$ is an eigenvalue of $AB$. True or false? Give reason.

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As David said, this is false. But since you do know that the product of the eigenvalues of a square matrix is the determinant, the fact that the determinant is multiplicative gives you some information about the original eigenvalues. –  Eric Gregor Apr 11 '12 at 18:13
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up vote 8 down vote accepted

False. Take $A=[ {1\atop 0}{0\atop 0}]$ and $B=[ {0\atop 0}{0\atop 1}]$. $\lambda=1$ is an eigenvalue of both $A$ and $B$, but $AB$ is the zero matrix.

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False. In addition to David's explicit counter-example, here's a naive counting argument: There are (typically) $n^2$ different products $\lambda\mu$, but only $n$ eigenvalues of $AB$.

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