# Is it possible to prescribe matrix monomial equations for polynomial equation systems in element's field?

It's me again. This time I got an idea from this question regarding polynomial equation systems arising from a matrix equation. Having read sporadically about polynomial equation systems and learned that the field to solve them is extremely theoretically challenging (algebraic geometry), I could not help but think about if one could go the other way around.

To try and design matrices in such a way that a monomial equation of the matrix would be the same as a polynomial equation system in the elements' field and then use a switch to some canonical basis ( ideally a diagonalization, but probably a block diagonalization could help quite a bit ). This would be useful for me as I have done much matrices and linear algebra.

1. Would this be useful or is it only boring polynomial equation systems we would be able to express?
2. If this is investigated, do you know any place to read more about how to do this?

## 1 Answer

PART 1. We want to solve the equation (F): $p(X)=A$, in the unknown $X\in M_n(K)$, where $K$ is a field, $p\in K[x]$ and $A\in M_n(K)$. For the sake of simplicity, assume that $A$ is diagonalizable over $K$. Note that, if $X$ is a solution, then $XA=AX$. Then we may assume that $A=\lambda I$ and (F) is in the form $q(X)=0_n$.

PART 2. We want to solve the equation (E): $p(X)=0_n$ - where $p\in K[x]$ - in the unknown $X\in M_n(K)$.

i) $K$ is an algebraic closed field. The roots of $p$ are $(\alpha_i)_{i\leq s}$ with multiplicity $(r_i)_{i\leq s}$. Let $J_r$ be the nilpotent Jordan block of dimension $r$. Then $X$ is a solution of (E) IFF $X$ is similar to $diag(U_1,\cdots,U_k)$ for any choice of $k$ and of the dimension $n_j$ of $U_j$ satisfying $n_1+\cdots+n_k=n$ and for any choice of $U_j$ among the matrices of following forms: $\alpha_i I_{n_j}+J_{n_j}$ with $i\leq s$ and $n_j\leq r_i$.

ii) $K$ is a field not alg. closed. We factor $p$ in irreducible over $K$: $p=p_1^{r_1}\cdots p_s^{r_s}$. The result is similar to that of i). It suffices to choose $U_j$ among the companion matrices of ${p_i}^{q}$ where $q=n_j/degree(p_i)$ and $q\leq r_i$.

iii) $K$ is an euclidean ring (for example $\mathbb{Z}$). This case is much more difficult. We must seek a number of ideal classes ; using Magma software, we can do that but, morover, we must find one representant in each class, that is not obvious. For example, try to solve $A^3=I_n$ where $A\in M_n(\mathbb{Z})$.