I am dealing with the following problem.
For $k,d \in \mathbb{N}$ we consider a set of square matrices $A =\{A_0,\ldots,A_{d-1}\} \subset \mathbb{C}^{k \times k}$ and associate with it a matrix polynomial as, $P_A(\lambda) = \sum_{i=0}^{d-1} A_i \lambda^{i}$, where $\lambda \in \mathbb{C}$. The eigenvalues of $P_A(\lambda)$ are defined as $\lambda(A) = \{\lambda : \text{det}(P_A(\lambda)) = 0\}$.
Now, consider the invertible matrix $M = (m_{i,j})_{{i,j} \in \{0,\ldots,d-1\}} \in \mathbb{C}^{d \times d}$, which defines a new set of matrices $B = (B_0,\ldots,B_{d-1})$ via $B_i = \sum_{j=0}^{d-1} m_{i,j} A_{j}$.
I am interested in the following question:
What is the set of invertible matrices $M$ such that $\lambda(A) = \lambda(B)$?
Some remarks:
- $d = 1$: the problem is trivival.
- $d = 2$: we can use the theory of matrix pencils to completely solve the problem (see e.g. the classic book of Gantmacher).
- $d>2$: A complete solution of the problem appears to be elusive. Can we get strong necessary conditions that $M$ has to fulfill?
Possible approaches for $d>2$:
To a matrix polynomial $P_A(\lambda)$ one can assign a compagnion matrix pencil $\tilde{P}_A(\lambda) = \tilde{A}_0 + \lambda \tilde{A}_1$, where $\tilde{A}_i \in \mathbb{C}^{k' \times k'}$,$k' = (d-1)k$ (see e.g. here for the precise definition). The eigenvalues of $P_A(\lambda)$ can then be determined from the eigenvalues of $\tilde{P}_A(\lambda)$. This might give some strong necessary conditions that $M$ has to fulfill.
There are various bound on the modulus of the elements in $\lambda(A)$ (see e.g. Bounds 1, Bounds). But it is not clear to me how they will be useful.
Any help is greatly appreciated.
Best