# Eigenvalues of matrices.

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.

-
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. –  user21725 Apr 11 '12 at 18:13
Welcome to math.SE: since you are fairly new, I wanted to let you know a few things about the site. In order to get the best possible answers, it is helpful if you say in what context you encountered the problem, and what your thoughts on it are so far; this will prevent people from telling you things you already know, and help them write their answers at an appropriate level. If this is homework, please add the [homework] tag; people will still help, so don't worry. Also, many find the use of imperative ("Find", "Show") to be rude when asking for help; please consider rewriting your post. –  Arturo Magidin Apr 12 '12 at 2:26
@ArturoMagidin Thank you for your guideline. I'll keep these things in mind. By the way, this is not homework. I took admission in an open university and I am learning mathematics all by myself. There is no one around me who can teach me a single concept. That's why sometimes I asked simple questions which I encountered in my learning process. –  Faisal Apr 12 '12 at 9:42

## 2 Answers

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.

-

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 $AB$.

-