Consider a monic quadratic matrix polynomial:

$$M(\lambda) = \lambda^2 I + \lambda A + B$$

Where $A$ is a real diagonal matrix, $B$ is a real positive semidefinite matrix (symmetric).

We are looking for the solution of the quadratic eigenvalue problem that it is the set of eigenvalues $\lambda$ and nontrivial eigenvectors $\psi$ such that $M(\lambda)\psi=0$.

In a particular case of $A = a I$, where $a \in \mathbb{R}$, the spectrum of $M (\lambda) $ could be described in terms of the spectrum of $B$ in the following way. Eigenvectors are the same and eigenvalues of $M(\lambda) $ are the solution of $$ \lambda^2 + a\lambda + d = 0$$ Where $d$ - eigenvalue of $B$.

Is there any connection between the spectrum of $M(\lambda)$ and the spectrum of $B$ with any diagonal $A$?


Because $B$ is symmetric it is diagonalizable by orthogonal eigenvectors $B = Q D Q^T$. Then the original polynomial could be represented as the following $$Q^T M(\lambda) Q = \lambda^2I + \lambda Q^T AQ + D$$

It is clear that in the mentioned particular case of $A = aI$, the term $Q^T A Q = a I$ and the matrix $Q^T M(\lambda) Q$ becomes diagonal and $\text{det}[Q^T M(\lambda) Q] = \prod_j \lambda^2 + a\lambda + d_j$.

Is it possible to exploit the fact that $A$ is diagonal to simplify $\text{det}[Q^T M(\lambda) Q]$?

  • $\begingroup$ I don't think it is possible to write them explicitly in terms of entries of matrices, but it should be possible to write their characteristic polynomial in terms of entries of matrices. $\endgroup$ – Sungjin Kim Feb 9 '19 at 20:08
  • $\begingroup$ It seems that you are not looking for the eigenvalues of $M(\lambda)$, but you are looking for roots of the polynomial equation $\det M(\lambda)=0$. They are different things. $\endgroup$ – Sungjin Kim Feb 10 '19 at 15:59
  • $\begingroup$ @i707107 probably there is an abuse of terminology. I am looking for the solution of the quadratic polynomial problem $M(\lambda)\psi=0$ (en.wikipedia.org/wiki/Quadratic_eigenvalue_problem). In this case eigenvalues and roots of a polynomial equation $\det M(\lambda)=0$ are the same. $\endgroup$ – Andrey Gorbunov Feb 10 '19 at 16:41
  • $\begingroup$ When you use $\phi_i^T A \phi_j=0$, are you assuming that $B$ can be diagonalized by orthogonal eigenvectors? $\endgroup$ – Sungjin Kim Feb 10 '19 at 22:08
  • $\begingroup$ @i707107 yes, I am because $B$ is semidefinite and, consequently, symmetric. $\endgroup$ – Andrey Gorbunov Feb 10 '19 at 23:24

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