Proving the relationship between left and right eigenvalues for a complex generalized eigenproblem

Question

A generalized right eigenproblem is defined as:

$A q = \lambda B q$

where $A$ and $B$ are complex matrices in general and $q$ represents the right eigenvector, which is a column vector, and $\lambda$ is the scalar right eigenvalue.

The generalised left eigenproblem for the same $A$ and $B$ matrices is:

$v A = \kappa v B$

where $v$ represents the left eigenvector, which is a row vector, and $\kappa$ is the scalar left eigenvalue. Let's assume that $\kappa \neq \lambda$ in general for now.

How do we prove that $\kappa = \lambda^*$, where the asterisk represents the complex conjugate operation?

I can construct one way:

1. Left-multiply the right eigenproblem with $v$ to make: $vAq = \lambda v B q$.
2. Right-multiply the left eigenproblem with $q$ to make: $vAq = \kappa vBq$.
3. By comparing the two resultant equations, we conclude that either $\lambda = \kappa$ or $vAq = 0$.

Clearly, this conclusion is not correct if $A, B, v, q$ can be complex in general.

Another way might be to use the conjugate transpose operation denoted by $H$.

1. Take the conjugate transpose of the left eigenproblem to make: $A^H v^H = \kappa^* B^H v^H$
2. ...?