# Generalized eigenvalue problem of Hermitian matrix (exist complex eigenvalues)

Let $\mathbf{A}, \mathbf{B} \in \mathbb{C}^{n \times n}$ be Hermitian and invertible. Show that exists eigenvalues $\lambda$ which satisfy $\mathbf{Av} = \lambda \mathbf{Bv}$ are not real.

My solution and question:

From the property of Hermitian we know that all eigenvalues of Hermitian matrix are real. (Omit the proof)

Then $\forall x \in \mathbb{C}^n$ we have $x^HAx = x^HP^HDPx = (Px)^HD(Px) = y^HDy = \lambda_1|y_1|^2 + \cdots + \lambda_n |y_n|^2$ which is a real number, where $D$ is diagonal matrix and the second inequality is derived form eigendecomposition.

Back to the question, $Av = \lambda Bv \Rightarrow v^HAv = \lambda v^HBv$.

Since $\mathbf{A}$ and $\mathbf{B}$ are Hermitian, the quadratic form should real. Therefore, the $\lambda = \frac{v^HAv}{v^HBv}$ is real.

Is the original problem wrong? Could anyone help me out? Thanks in advance!

Your considerations are fine. The original problem wrong, as can also seen by taking $\mathbf{B} =\mathbf{I}$.