# Simultaneous triangularization of two diagonal matrices and a symmetric matrix

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".

In general, though, what you are asking seems too strong; I don't think one can solve a QEP easily in that way. The usual routes (linearize the problem using the companion form, see e.g. http://dx.doi.org/10.1137/S0036144500381988) should work better. There are several canonical forms for pairs of matrices (QZ factorization, simultaneous diagonalization by congruence), but working with three matrices at the same time is way more difficult, and I don't think that the fact that $M$ and $D$ commute helps here.
Shameless advertising: with a co-author, we have recently studied a problem arising from Markov chain modelling in which the matrices have that exact same structure (diagonal-diagonal-generator of a cont.time Markov chain): see https://arxiv.org/abs/1605.01482. The only difference with respect to your setup is that $D$ has mixed signs (but maybe a change of variables in $\lambda$ can take care of that). I don't know what is the origin of your quadratic eigenvalue problem and what your computing goal is, but it might be related. What we do there is, essentially, finding the subspace spanned by the eigenvectors with $\lambda_i \leq 0$, or finding the general stable solution of the differential equation $M\ddot x + D \dot x + Jx=0$.
• @ZPascal I am not sure what you want to prove. If you want to show that the determinant of that matrix polynomial is in the form $\Pi (\lambda^2M_i + \lambda D_i + \mu_i) = 0$, with $M_i$, $D_i$ the diagonal entries of the original matrices $M$ and $D$, I am afraid that it is false. If it were true, it would imply that the eigenvalues can be divided in pairs with sum $-D_i/M_i$, which should be false in general. If you drop the requirement that $M_i$ and $D_i$ have to be the diagonal entries of $M$ and $D$, then any degree-$2n$ polynomial can be written as a product of $n$ degree-$2$ ones. – Federico Poloni Apr 22 '17 at 18:04