on existence of matrices X, Y s.t. XAY is diagonal over non-commutative ring - MathOverflow

on existence of matrices X, Y s.t. XAY is diagonal over non-commutative ring

2010-11-30T15:58:18Z

Given $A\in Mat_{n\times n}(R)$ where $R$ is a non-commutative associative ring are there exist any (non-zero) matrices $X, Y\in Mat_{n\times n}(R)$ such that $XAY=diag(a_1, \ldots , a_n)$ for some $a_i$? The interesting answer for me is if $A=(x_{i,j})$ and $R=\mathbb Z [x_{i,j}]$ (free associative non-commutative algebra on $x_{i,j}$ over $\mathbb Z$).

For example if $R$ is commutative then we put $X=Id$, $Y=(det(A_{i,j}))$ and get $XAY=det A\cdot Id$. What about non-commutative polynomials?

Upd1: I want to have a non-commutative polynomial equality. Also I want $X$ and $Y$ to be in general invertible.

Upd2: Ok, I've understood that Update1 wasn't correct. I'm interested in having such matrices over $R=Mat_{m\times m}(A)$ where $A$ is a commutative ring with $1$.

Answer by Mark Sapir for on existence of matrices X, Y s.t. XAY is diagonal over non-commutative ring

2010-11-30T17:10:43Z

Take $X=Y=E_{1,1}$ (the matrix unit). Then $XAY=x_{1,1}E_{1,1}$, a diagonal matrix. If $R$ does not have 1, take $X=Y=aE_{1,1}$ for any $a\ne 0\in R$. Then $XAY=ax_{1,1}aE_{1,1}$ (it may be a zero matrix, but zero matrix is diagonal).

Update. Since you now want to find invertible $X,Y$, I would recommend starting with $2\times 2$-matrices and reading the book by Cohn, "Free rings and their relations", especially Chapter 2, Section 2.6.

Answer by zroslav for on existence of matrices X, Y s.t. XAY is diagonal over non-commutative ring

2010-12-09T19:33:24Z

I've understood that if $R=Mat_n(K)$ then for every $A\in Mat_m(R)$ exist $B\in Mat_m(R)$, s.t. $AB=\lambda Id$. $A=(a_{ij,kl})$ is a $mn\times mn$-matrix over $K$ and $B=A^{V}$ is a $m\times m$-matrix over $R$.