eigenproblem and characteristic equation Eigenvalues $\lambda$ of a matrix $A$ are defined as $Ax = \lambda x$, where $x$ are the eigenvectors. The characteristic polynomial is $\det(A-\lambda I)$.
Say I have the expression $Ax = \lambda Bx$, where $B$ is a matrix as well. Does this equation have a characteristic polynomial that enables me to find its eigenvalue $\lambda$?
 A: Suppose B is invertible, so $B^{-1}$ exists. This means you can re-write your equation to:
$\begin{align}Ax &= \lambda B x\\
\Leftrightarrow Ax &= B \lambda x\\
\Leftrightarrow \underbrace{B^{-1} A}_{=: C} x &= \lambda x\\
\Leftrightarrow Cx &= \lambda x
\end{align}$
So you end up in the same situation as before.
A: You are always allowed to do the generalization proposed by Isaac, which remains valid, even if $B$ is not invertible.
Namely, there is a non-zero vector $x$ such that 
$$
Ax = \lambda Bx \qquad \Longleftrightarrow \qquad (A - \lambda B)x = 0 \ .
$$
Which means that the matrix $A - \lambda B$ has a non-trivial null space (for instance, $x$ belongs to it). Which means its determinant must be zero and vice-versa: if $\det (A-\lambda B) = 0$, then the matrix $A - \lambda B$ has non-trivial null space. 
All in all, 
$$
\text{There is  $\ x \neq 0\ $ such that} \ \ Ax = \lambda Bx \qquad \Longleftrightarrow \qquad \det (A - \lambda B) = 0 \ .
$$
And I guess you can call $\det (A - \lambda B)$ the characteristic polynomial of $A$ and $B$, or something like that if you whish.
EDIT. If the matrix $B$ is invertible, then $\det (B) $ and $ \det (B^{-1}) = 1/\det(B) \neq 0$ and the last equality is equivalent to
$$
0 = \det (B^{-1}) \det (A - \lambda B) = \det (B^{-1} A - \lambda B^{-1}B ) = \det (B^{-1} A - \lambda I ) \ .
$$
