I am wondering about the difference between simultaneous diagonalization and the general eigenvalue problem. Here is my understanding of simultaneous diagonalization and the general eigenvalue problem:
Simultaneous Diagonalization: Two matrices, $A$ and $B$, are said to be simultaneously diagonalizable if and only if there exists an invertible matrix $P$ such that $P^{-1}AP$ is diagonal and $P^{-1}BP$ is diagonal.
Generalized Eigenvalue Problem: The set of vectors, $V$, that satisfy $AV = \Lambda BV$ are the generalized eigenvectors. This results in $V'AV = \Lambda$ and $V'BV = I$.
The reason I am confused is because the generalized eigenvalue problem results in a matrix that diagonalizes both $A$ and $B$. However, I believe that most of the time it would not satisfy the constraints from the definition of simultaneously diagonalization, $P^{-1}AP$ is diagonal and $P^{-1}BP$ is diagonal, because $V$ is not guaranteed to be unitary, so $V^{-1}$ is not equal to $V'$ and $V'AV$ is not equal to $V^{-1}AV$. The only time they are equal would be when $V$ is unitary which I think is only when $B^{-1}A$ is Hermitian.
Edit Just a quick follow up as to why I believe $V$ is unitary only when $B^{-1}A$ is Hermitian.
As shown in this tutorial on page 6, the generalized eigenvalue problem can be viewed as the eigenvalue problem for the matrix $C$ where $C=B^{-1}A$. By the spectral theorem, C has a unitary eigenvalue decomposition when it is Hermitian.
Is this understanding correct? Thank you for the help.