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 $1$.

  • $\begingroup$ If the ring has a unity, then it is easy: just choose $a_{ij} \neq 0$ then $$ e_{1i}Ae_{j1}=a_{ij}e_{11} $$ where $e_{kl}$ is the matrix with $1$ in the $(k,l)$ entry and zero elsewhere. If the ring does have $1$, the statement is not generally true, since you can define $xy=0$ for all $x,y \in R$. $\endgroup$ Nov 30, 2010 at 17:11
  • $\begingroup$ You want $X$ and $Y$ invertible or at least non-zero divisors. Otherwise the answer is no for obvious reasons. $\endgroup$ Nov 30, 2010 at 19:12
  • $\begingroup$ @Andreas: I want to have a generalization of the equality: $A A^{V}=det A\cdot Id$. $\endgroup$
    – zroslav
    Nov 30, 2010 at 19:31

2 Answers 2


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.


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 $R$.


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