I have the following quadratic eigenvalue problem:
$\det(\lambda^2M + \lambda D + J)=0$
where, M and D are both $n \times n$ real diagonal matrix with positive diagonal entries; $J$ is
1) a nearly symmetric n-by-n real matrix.
2) One of the eigenvalues of $J$ is guaranteed to be zero because each row sum of $J$ is guaranteed to be zero.($J$ looks similar to the so-called "Laplacian matrix" in Graph theory).
3) $J$ is positive semi definite,i.e. the other eigenvalues are guaranteed to be larger than zero.
I want to find a decoupled analytic solution for the expression of each eigenvalue. I thought about Schur decomposition (or, "Schur Unitary Triangularization"), i.e. whether $\exists \text{ a unitary } U $, such that $U^HJU = T$, where $T$ is a upper triangular matrix with diagonal elements as the same as the eigenvalues of $J$.
Previously, I took for granted that $U^HMU$ and $U^HDU$ are still diagonal; however this can be wrong; thus, now I am thinking about whether there is any condition such that $U^HMU$ and $U^HDU$ are also upper triangular.
If this can be achieved, then the determinant can be reduced to upper triangular structure and its value is: $\Pi (\lambda^2M_i + \lambda D_i + \mu_i) = 0$, where $\mu_i$ is the eigenvalue of $J$, and each $\lambda_i$ can be solved for from the well-known root formula of the quadratic equation.
I spent a whole day reading the literature and found so far some of the "best" "Simultaneously triangularization" sufficient conditions are from
"Simultaneous Triangularization"(Book) By Heydar Radjavi, Peter Rosenthal, pp. 19-21 Theorem 1.6.4 and 1.6.6. (Google provides a preview of this part.)
Although the theorem itself is a constructive one, it is not very straightforward to use. And it looks like the book's content is hard for me, trying to use "representation theory" and/or "Lie algebras" to prove things, which is far beyond my current level.
So, is there any other straightforward method to reach my goal? What if I loosen the requirement for J, say J is strictly symmetric?
Note: Another approach I am trying to look at now is "the quadratic eigenvalue problem".