1

Consider the generalized eigenvalue problem $$A\textbf{v}=\lambda B \textbf{v},$$ where $A$ is assumed to be symmetric, nonsingular with distinct eigenvalues and $B$ symmetric and positive definite and let $(\lambda_i,\textbf{v}_i)$ be the $i$-th eigenpair of the problem. I would like to show that the $\textbf{v}_i$ can be chosen such that $$\textbf{v}_i^TB\textbf{v}_j=\delta_{ij}.$$ I have managed to show that for $i \neq j$ the vectors $\textbf{v}_i$ and $\textbf{v}_j$ are orthogonal with respect to $B$ (i.e. $\textbf{v}_i^TB\textbf{v}_j= 0$ for $i \neq j$), but I am not sure how to show that in the case $i=j$ I can have $$\textbf{v}_i^TB\textbf{v}_i=1.$$

wwinters57
  • 745
  • 1
    $B$ is positive definite, therefore $w^T Bw>0$ for any $w\neq 0$. Thus if $w_i$ is some eigenvector ($Aw_i = \lambda_i Bw_i)$, then define $v_i:= w_i/(w_i^T Bw_i)^{1/2} $, and so $v_i^T B v_i=1$. – Mick Mar 31 '20 at 13:07

1 Answers1

1

I'm assuming your matrices are real.

The generalized eigenvectors $v$ correspond to eigenvectors $w$ of $B^{-1/2} A B^{-1/2}$ by $v = B^{1/2} w$. Since $B^{-1/2} A B^{-1/2}$ is symmetric, it has an orthonormal basis of eigenvectors $w_j$. Then $v_i^T B v_j = w_i^T w_j = \delta_{ij}$.

Robert Israel
  • 470,583
  • It might be worth noting that transforming the set of w vectors back to the original problem space v will generally destroy their orthogonality. (unless B is unitary) – Mark Borgerding Jan 31 '23 at 20:14