Algebra - Decomposition of a matrix polynomial - MathOverflow most recent 30 from http://mathoverflow.net 2013-05-20T04:51:38Z http://mathoverflow.net/feeds/question/83254 http://www.creativecommons.org/licenses/by-nc/2.5/rdf http://mathoverflow.net/questions/83254/algebra-decomposition-of-a-matrix-polynomial Algebra - Decomposition of a matrix polynomial Federico Carlini 2011-12-12T16:36:35Z 2012-10-05T09:33:20Z <p>Dear All,</p> <p>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. </p> <p>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]\$?</p> <p>\$ax = xa\$ iff \$a\$ is in the commutator of \$E\$. Moreover, \$ A\$ is unitary. </p> <p>Thanks a lot,</p> <p>Federico Carlini</p> http://mathoverflow.net/questions/83254/algebra-decomposition-of-a-matrix-polynomial/83259#83259 Answer by Alfonso Gracia-Saz for Algebra - Decomposition of a matrix polynomial Alfonso Gracia-Saz 2011-12-12T17:16:58Z 2011-12-12T17:16:58Z <p>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.</p> http://mathoverflow.net/questions/83254/algebra-decomposition-of-a-matrix-polynomial/83263#83263 Answer by Federico Carlini for Algebra - Decomposition of a matrix polynomial Federico Carlini 2011-12-12T17:32:46Z 2011-12-12T17:32:46Z <p>Dear Alfonso,</p> <p>I'm not understanding why \$A\$ has to be a domain.</p> http://mathoverflow.net/questions/83254/algebra-decomposition-of-a-matrix-polynomial/83648#83648 Answer by zroslav for Algebra - Decomposition of a matrix polynomial zroslav 2011-12-16T18:13:32Z 2011-12-16T18:13:32Z <p>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: <a href="http://en.wikipedia.org/wiki/Smith_normal_form" rel="nofollow">http://en.wikipedia.org/wiki/Smith_normal_form</a>). 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).</p>