linear algebra - 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 $\lambda$?

Answer

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

$$0 = \det (B^{-1}) \det (A - \lambda B) = \det (B^{-1} A - \lambda B^{-1}B ) = \det (B^{-1} A - \lambda I ) \ .$$