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Let $S\subseteq R$, where $R$ a commutative ring. Write $A\in S^{n\times n}(r)$ to mean that $A$ is an $n\times n$ matrix of rank $r$ with entries in $S$, with the rank understood in the sense of minors. Let $S^{n\times m}$ denote the set of all $n\times m$ matrices with entries in $S$.

Let us say that a matrix $A\in S^{n\times n}(r)$ has the full-rank factorization property (FRFP) over $S$ if $A$ can represented as the product $XY$ of matrices $X\in S^{n\times r}$ and $Y\in S^{r\times n}$.

If $R$ is a field, then it is easy to see that any $A\in R^{n\times n}(r)$ has the FRFP over $R$. E.g., let the columns of $X$ constitute a basis of the column space of $A$ and let the entries of $Y$ be the coefficients of the columns of $A$ in that basis.

Concerning the case $r=1$: It is not hard to see that any $A\in\N^{n\times n}(1)$ has the FRFP over $\N:=\{1,2,\dots\}$, as shown in Proposition 1. This result and its proof remain valid if $\N$ is replaced by $\Z\setminus\{0\}$ or $R\setminus\{0\}$, where $R$ is any unique factorization domain.

The case $r\ge2$ seems quite a bit more complicated in general; in what follows, let us indeed assume $r\ge2$. It is obvious that no matrix $A\in\N^{n\times n}(r)$ with at least one entry equal $1$ can have the FRFP over $\N$ -- because for all $X\in\N^{n\times r}$ and $Y\in\N^{r\times n}$ all the entries of $XY$ are $\ge2$. Also, for $\N_0:=\{0,1,\dots\}$, it is not hard to see that matrix $A=\left( \begin{array}{ccc} 3 & 2 & 0 \\ 5 & 4 & 4 \\ 0 & 1 & 6 \\ \end{array} \right)\in\N_0^{3\times3}$ does not have the FRFP over $\N_0$.

However, I have not been able to find a matrix $A\in\Z^{n\times n}(r)$ which does not have the FRFP over $\Z$.

So, a question is: Does such a matrix exist? More generally, are there good necessary and/or sufficient conditions on $S$ in order for all matrices $A\in S^{n\times n}(r)$ to have the FRFP?

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    $\begingroup$ When $S$ is a principal ideal domain (for example $S=\mathbb Z$) it seems to me that such a factorisation is given by Smith normal form. Am I missing something? $\endgroup$
    – Simon L Rydin Myerson
    Commented May 1, 2018 at 20:06
  • $\begingroup$ @SimonLRydinMyerson : Thank you. So, now I have learned about the Smith normal form -- had never heard of it before. $\endgroup$
    – Iosif Pinelis
    Commented May 1, 2018 at 21:28
  • $\begingroup$ @Iosef Pinelis glad I could help! When S is not at least a ring, this seems like a fairly difficult question. Was your main interest in the integer case? $\endgroup$
    – Simon L Rydin Myerson
    Commented May 3, 2018 at 8:47
  • $\begingroup$ @SimonLRydinMyerson : Yes, initially and principally, integers. The principal ideal domain case, in which the Smith normal form is guaranteed to exist, certainly seems rather general to me. Concerning proper subsets of rings, it seems the case r=1 is the only exception when something interesting of this sort may occur. $\endgroup$
    – Iosif Pinelis
    Commented May 3, 2018 at 12:53

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