$SV = TVD$ where $S$ is symmetric positive semi definite, $T$ is symmetric positive definite, $V$ is eigenvector matrix and $D$ is eigenvalue matrix.
In this case, can we guarantee $V$ is orthogonal? i.e. $V^{H} * V = $I?
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- $\begingroup$ Please remember to use MathJAX when posting equations. $\endgroup$ml0105– ml01052014-03-26 02:50:32 +00:00Commented Mar 26, 2014 at 2:50
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3 I think this is not true in general. You can however prove, that the eigenvectors are $T$- and $S$-orthogonal:
Let $v_i$, $v_j$ be eigenvectors to eigenvalues $\lambda_i\ne\lambda_j$. Then we have $$ v_i^HSv_j = \lambda_j v_i^HTv_j, \quad v_j^HSv_i = \lambda_i v_j^HTv_i. $$ Hence it holds, $$ \lambda_j v_i^HTv_j = \overline{\lambda_i v_j^HTv_i} = \lambda_i v_i^HTv_j, $$ which implies $$ v_i^HTv_j = 0. $$ If both $\lambda_i\ne0$, $\lambda_j\ne0$ then also $$ v_i^HSv_j = 0 $$ can be proven (with analogous arguments).
- $\begingroup$ Could the joint eigenvalues (i.e., the elements of $D$) related to the eigenvalues of $S$, $T$? Could it be more easy to find them in that particular case (where $S$, $T$ are symmetric and positive-definite)? Thanks! $\endgroup$nullgeppetto– nullgeppetto2015-06-15 18:44:49 +00:00Commented Jun 15, 2015 at 18:44
- $\begingroup$ Hard to say. The problem is that eigenvectors of $S$ are hardly eigenvectors of $D^{-1}S$ or $SD$. Maybe you could open a new question with your problem? $\endgroup$daw– daw2015-06-15 19:33:06 +00:00Commented Jun 15, 2015 at 19:33
- $\begingroup$ That's what I though! Thanks for your consideration anyway! In fact, I'm interested in a numerical solution (in C, specifically), so that's the way I asked. Thanks again! $\endgroup$nullgeppetto– nullgeppetto2015-06-15 19:39:44 +00:00Commented Jun 15, 2015 at 19:39