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