If $v$ is an eigenvalue of $A$ and $c$ is an eigenvalue of $B$, must $vc$ be an eigenvalue of $AB$? Let $v$ be an eigenvalue of $A$ and $c$ be an eigenvalue of $B$. Is the product of $v$ and $c$ equal to an eigenvalue of $AB$?
 A: Hint: There are matrices $A$ and $B$, each of which have non-zero eigenvalues, such that $AB=0$.
If you need a suggestion of such matrices, you can mouse over the gray area below.

 Suggestion: Let $A=\begin{pmatrix} 2 & 0 \\ 0 & 0 \end{pmatrix}$ and $B=\begin{pmatrix} 0 & 0 \\ 0 & 3 \end{pmatrix}$. 

A: $$
\begin{pmatrix}1 & 0\\0 & -1\end{pmatrix}
\begin{pmatrix}0 & 1\\1 & 0\end{pmatrix}=
\begin{pmatrix}0 & 1\\-1 & 0\end{pmatrix},
$$
but while $1$ and $-1$ are clearly eigenvalues of both the matrices on the left, the matrix on the right does not have any eigenvalues (over the real numbers).
A: Consider the following matrices
$$
A=\left( \matrix{1&-1\\0&0}\right)\qquad B=\left( \matrix{0&1\\0&1}\right)
$$
and seek a counterexample.
Note: I have no idea why I did not directly set all the off-diagonal coefficients to $0$. Do it yourself while I leave it like this for self-punishment, of for a (useless) non-diagonal example.
A: $1$ is an eigenvalue of $
        \begin{pmatrix}
        1 & 0\\
        0 & 2\\
        \end{pmatrix},$
$3$ is an eigenvalue of $
        \begin{pmatrix}
        2 & 0\\
        0 & 3\\
        \end{pmatrix}$ 
but $3$ is not an eigenvalue of $
        \begin{pmatrix}
        1 & 0\\
        0 & 2\\
        \end{pmatrix}
        \begin{pmatrix}
        2 & 0\\
        0 & 3\\
        \end{pmatrix}=
        \begin{pmatrix}
        2 & 0\\
        0 & 6\\
        \end{pmatrix}.$
A: Here is a simple theoretical reasoning that the answer is no:
Pick $A,B$ to be matrices, each of them having exactly $n$ distinct eigenvalues. Then the products of eigenvalues of $A,B$ can take $n^2$ distinct values, and it is easy to construct examples where the products are pairwise distinct. But $AB$ can only have $n$ eigenvalues.... 
