Algebra - Decomposition of a matrix polynomial - MathOverflow

Algebra - Decomposition of a matrix polynomial

2011-12-12T16:36:35Z

Dear All,

This is related with a problem that I'm trying to solve on my PhD dissertation in econometrics, and I thought that some mathmatician can know the answer.

What is known about a possible extension, $E$ , of the ring, $ A$ , of all n-by-n matrices with entries in $\mathbb{C}$ such that any non-constant polynomial of $ A[x] $ splits in a product of linear factors in $E[x]$?

$ax = xa$ iff $a$ is in the commutator of $E$. Moreover, $ A$ is unitary.

Thanks a lot,

Federico Carlini

Answer by Alfonso Gracia-Saz for Algebra - Decomposition of a matrix polynomial

2011-12-12T17:16:58Z

If you only need to show that $E$ exists, that is easy. $A$ is a domain, so it has a field of quotients $K$. Let $E$ be the algebraic closure of $K$. Then $E$ satisfies the condition you want. This is non-constructive, however.

Answer by Federico Carlini for Algebra - Decomposition of a matrix polynomial

2011-12-12T17:32:46Z

Dear Alfonso,

I'm not understanding why $A$ has to be a domain.

Answer by zroslav for Algebra - Decomposition of a matrix polynomial

2011-12-16T18:13:32Z

This is true because of following: any matrix over principal ideal domain can be reduced by elementary Gauss transformations to the Smith normal form (see this: http://en.wikipedia.org/wiki/Smith_normal_form). The ring of polynomials over any field is such a domain. Smith normal form can be splitted to linear factors. Elementary Gauss transformations may be of the form $(z, w)\mapsto (z+p(x)w, w)$, where $p(x)$ is not linear. I think that these transformations cannot be splitted in linear factors (I don't know how to prove it - please let me know if you do know this, I'm curious about).