Properties of generalized eigenvalue problem when hermitian

Suppose $Av_1=\lambda_1Bv_1,Av_2=\lambda_2Bv_2$.

Then $\lambda_1v_2^*Bv_1=v_2^*Av_1=v_2^*A^*v_1=(v_1^*Av_2)^*=(\lambda_2v_1^*Bv_2)^*=\lambda_2v_2^*B^*v_1=\lambda_2v_2^*Bv_1$

As $\lambda_1,\lambda_2$ are real and distinct, $v_2^*Bv_1=0$.

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 $(L^* v_1)^* (L^* v_2) = v_1^* L L^* v_2 = v_1^* B v_2 = 0$.