# The spectrum of a quadratic matrix polynomial

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 $$A$$?

## Update

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 $$\text{det}[Q^T M(\lambda) Q]$$?

• 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. – Sungjin Kim Feb 9 '19 at 20:08
• 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. – Sungjin Kim Feb 10 '19 at 15:59
• @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. – Andrey Gorbunov Feb 10 '19 at 16:41
• When you use $\phi_i^T A \phi_j=0$, are you assuming that $B$ can be diagonalized by orthogonal eigenvectors? – Sungjin Kim Feb 10 '19 at 22:08
• @i707107 yes, I am because $B$ is semidefinite and, consequently, symmetric. – Andrey Gorbunov Feb 10 '19 at 23:24