Consider $Z = K - i \omega S - \omega^2 M$, where: $\omega$ is a positive real number; $K$ is a real, symmetric, and positive semi-definite matrix; $M$ is a real, symmetric, and positive definite matrix.

$S$ is a non-zero matrix.

Which property (or properties) $S$ should satisfy so that $Z$ is invertible? Or so that $(Z, M)$ has an eigendecomposition $Z = M \Phi \Lambda \Phi^T M$ (with $\Phi^T M \Phi = I$)?

Or which reference book discusses this class of problems?

Remark 1: Matrices like $Z$ arise in the discretization of the Helmholtz equation with absorbing boundary conditions or of the frequency response analysis with damping.

Remark 2: Many posts on the forum discuss the matrix $\left[ \begin{array}{cc} 1 & i \\ i & -1 \end{array} \right]$ as an example of non-diagonalizable complex symmetric matrix. Here I want to know which non-zero matrix $S$ would work.

EDIT: Generalization of proportional damping gives us a family of $S$ matrices that would allow the eigendecomposition: $S = M \sum_{j = J_1}^{J_2} s_j \left( M^{-1} K \right)^j$

  • $\begingroup$ Can $K$ or $M$ be zero? $\endgroup$ – Leo Jul 30 '19 at 20:44
  • $\begingroup$ @Leo $K$ or $M$ can not be zero. Eventually, we could consider: when does $W = K - i \omega S$ have an eigendecomposition $W = \Psi \Theta \Psi^T$ with $\Psi^T \Psi = I$ ? $\endgroup$ – user7440 Jul 30 '19 at 21:27
  • $\begingroup$ We can reduce this to the case in which $M = I$ by noting that $$ M^{-1/2}ZM^{-1/2} = \tilde K - i \omega \tilde S - \omega^2 I $$ where $\tilde K = M^{-1/2} K M^{-1/2}$ is positive semidefinite. We can then further reduce your problem to the case where $\tilde K$ is diagonal. $\endgroup$ – Ben Grossmann Jul 30 '19 at 22:00
  • $\begingroup$ @Omnomnomnom I agree. $\endgroup$ – user7440 Jul 30 '19 at 22:16

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Browse other questions tagged or ask your own question.