What kind of Eigenvalue Problem is this? I want to solve this kind of eigenvalue problem
$$\Bbb{A}(\omega) \vec{T} = k\Bbb{B}(k)\vec{T}$$
where $\Bbb{A}$ and $\Bbb{B}$ are both nonsymmetric, complex matrices. $\omega$ is real, and the eigenvalue $k$ is complex. $\vec{T}$ is the eigenvector. Note that $\Bbb{B}$ is dependant on $k$.
 A: The condition is equivalent to
$$
\begin{align}
&\Bbb{A}(\omega) \vec{T} - k\Bbb{B}(k)\vec{T} = 0 \\
\iff &[\Bbb{A}(\omega)- k\Bbb{B}(k)]\vec{T} = 0.
\end{align}
$$
This happens if the kernel $\textrm{ker}(\Bbb{A}(\omega)- k\Bbb{B}(k))$ is non-empty, hence we must have
$$
\det(\Bbb{A}(\omega)- k\Bbb{B}(k)) = 0
$$
which you can solve numerically or analytically depending on your application.
Simple example
Let's take $\Bbb A(\omega)$ to be the rotation of angle $\omega$ and
$$
\Bbb B(z) = \begin{pmatrix}0 & 0 \\ -z & 0 \end{pmatrix}.
$$
The equation to solve is
$$
\det\begin{pmatrix} \cos \omega & - \sin \omega \\ 
                    \sin \omega + z^2&  \cos \omega
     \end{pmatrix} = 0,
$$
which becomes
$$
\begin{align}
\cos^2 \omega - (z^2 + \sin \omega)\sin \omega = 0 \\
\iff z^2 + \sin \omega = \cot(\omega)\cos(\omega) \\
\iff z^2 =  \cot(\omega)\cos(\omega) - \sin \omega \\
\iff z^2 = \frac{\cos 2\omega}{\sin \omega}
\end{align}
$$
For $\omega = \pi/6$ we get $z^2 = 1$ and you can then see by inspection that
$$
\begin{pmatrix}\sqrt 3 /2 & 1/2 \\
                3/2  & \sqrt 3 / 2
 \end{pmatrix}
$$
has 
$$
\textrm{ker}(\Bbb{A}(\omega)- k\Bbb{B}(k)) = \{(-w, \sqrt{3}w ) : w\in \mathbb C\}
$$
which in turn implies that $T=(-1,\sqrt{3})$ satisfies
$$
\Bbb A(\pi/6) T = \Bbb B(1) T
$$
but also
$$
\Bbb A(\pi/6) T = -\Bbb B(-1) T.
$$
