TITLE: Properties of generalized eigenvalue problem when hermitian
QUESTION [6 upvotes]: This Wikipedia page says that, for the generalized eigenvalue problem 
$$\boldsymbol{A}\boldsymbol{v}=\lambda\boldsymbol{B}\boldsymbol{v},$$
if $\boldsymbol{A}$ and $\boldsymbol{B}$ are hermitian and $\boldsymbol{B}$ is positive-definite, then (1) eigenvalues $\lambda$ are real; (2) eigenvectors $\boldsymbol{v}_1$ and $\boldsymbol{v}_2$ with distinct eigenvalues are $\boldsymbol{B}$-orthogonal ($\boldsymbol{v}_1^*\boldsymbol{B}\boldsymbol{v}_2=0$).
How to prove (2)? I found the proof of (1) like this, but I can't find the proof of (2). The reference of this property on the Wikipedia page doesn't give the proof either.

REPLY [3 votes]: These properties follow from the properties of Hermitian matrices.
Let $B=LL^*$ be the Cholesky decomposition of $B$, then if $Ax= \lambda L L^* x$, we have $L^{-1} A L^{-*} y = \lambda y$ where  $y = L^* x$.
For (1) we see that the eigenvalues of the pencil are the eigenvalues of $L^{-1} A L^{-*}$, hence real.
For (2) suppose $\lambda_k, v_k$ are an eigenvalue, eigenvector pair of the pencil (with distinct eigenvalues),
then $\lambda_k, L^*v_k$ are an eigenvalue, eigenvector pair of $L^{-1} A L^{-*}$
and hence $(L^* v_1)^* (L^* v_2) = v_1^* L L^* v_2 = v_1^* B v_2 = 0$.