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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.

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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, $v_2^*Bv_1=0$.

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

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  • $\begingroup$ Thanks a lot for your answer! Btw I guess there's a typo in the last formula which I think should be $v_1^*LL^*v_2$ $\endgroup$ – WuQ Feb 3 '18 at 6:11
  • $\begingroup$ Thanks. Fixed it :-). $\endgroup$ – copper.hat Feb 3 '18 at 6:13

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