When does $A\mathbf{v} = \lambda B\mathbf{v}$ admit a basis of solutions?

Question

Let $A, B \in \mathbb{C}^{n \times n}$ be Hermitian matrices, and consider the so-called generalized eigenvalue problem $$A\mathbf{v} = \lambda B\mathbf{v}$$ where $\lambda \in \mathbb{C}$ is called a generalized eigenvalue of $A$ with respect to $B$, and $\mathbf{v} \in \mathbb{C}^n$ is the corresponding generalized eigenvector. It seems to be a well-known fact that if $B$ is positive-definite, then this problem admits a family of solutions $(\lambda_1, \mathbf{v}_1), \dots, (\lambda_n, \mathbf{v}_n)$ with $\lambda_1, \dots, \lambda_n \in \mathbb{R}$ and $\mathbf{v}_1, \dots, \mathbf{v}_n$ linearly independent, but I've never actually seen a proof. (See, for example, this Wikipedia article, whose references are unfortunately unavailable to me.) Does anyone know how the argument goes?

Answer 1

Hints:

Note that $A \sqrt{B}^{-1} \sqrt{B} v = \lambda \sqrt{B} \sqrt{B}v$ and so $\sqrt{B}^{-1} A \sqrt{B}^{-1} \sqrt{B} v = \lambda \sqrt{B}v$.

The matrix $\sqrt{B}^{-1} A \sqrt{B}^{-1}$ is Hermitian.