Number of solutions of a matrix algebraic equation
Reading that, I found that:
given a matrix algebraic equation $$ X^n+A_1X^{n-1}+\cdots + A_n=0 $$ where the cofficients $A_1\cdots,A_n$ as well as solutions $X$ are supposed to be square complex matrices of some order $k$....... generically has $\binom {nk}{k}$ rather $n$ solutions.
Someone know how can we prove this statement, or where I can find a proof?
Have a look at pages 2,4,5 (Theorem 1) of my following paper (free advertising) http://arxiv.org/abs/1304.2506
A slightly different version is submitted to LAA.
Answer to Emilio. 1. In fact, an old version (v2) is posted on arXiv. A recent one (v3) will be readable from March 3.
A simple solution has multiplicity $1$. (cf. def. 3,ex. 1,def. 4,ex. 2).
The considered result is valid when $A_1,\cdots,A_n$ are generic matrices (cf. def. 1). For instance, if you randomly choose the $(A_i)_i$, then (except if you are very unlucky), you obtain the required number of simple solutions.
About "... the matrices $X_h$ that are solutions of the equation (1) are ALL the matrices that have as eigenvalues...". "ALL" is incorrect: we choose $k$ values amongst the $nk$ roots of (2); then we construct the unique solution that admits these $k$ values as eigenvalues. In particular the complexity of the problem (1) is the complexity of the resolution of(2).
The answer to this OP is in the paper cited by @loup blanc. I added this post only to complete the initial question with an answer easily accessible to readers that sums up the main result in the paper.
It seems that the statement that an equation of the form: $$ (1) \qquad A_nX^n+A_{n-1}X^{n-1}+\cdots + A_0=0 $$ (where $A_i$ and $X$ are square complex matrices of order $k$) has $\binom {nk}{k}$ solutions, was found by J.J. Sylvester, but was communicated without a proof.
The proof given in the cited paper shows that the matrices $X_h$ that are solutions of the equation $(1)$ are matrices that have as eigenvalues $k$ numbers $\{\lambda_i\}\quad (\lambda_1 \in \{1,\cdots,nk\})$ , where the $\lambda_i$ are solutions of the equation: $$ (2) \qquad \mbox{det}\left(\lambda^n A_n+ \cdots+\lambda A_1+A_0\right)=0 $$ that is obviously an equation of degree $nk$, so the $X_h$ are exactly $\binom {nk}{k}$.
In the paper it's also proved that the Galois group of $(2)$ is $S_{nk}$ so the entries of the solution of $(1)$ can be calculated by radicals if and only if $n=k=2$.