# Invertibility (and eigen-decomposition) of complex symmetric matrix?

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: $$S = M \sum_{j = J_1}^{J_2} s_j \left( M^{-1} K \right)^j$$

• Can $K$ or $M$ be zero? – Leo Jul 30 '19 at 20:44
• @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 $\Psi^T \Psi = I$ ? – user7440 Jul 30 '19 at 21:27
• 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. – Ben Grossmann Jul 30 '19 at 22:00
• @Omnomnomnom I agree. – user7440 Jul 30 '19 at 22:16