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This is an attempt at an analogy with prime numbers. Let's consider only square matrices with positive integer entries. Which of them are 'prime' and how to decompose such a matrix in general?

To illustrate, there is a product of two general $2 \times 2$ matrices:

$$AB=\left[ \begin{matrix} a_{11} & a_{12} \\ a_{21} & a_{22} \end{matrix} \right] \left[ \begin{matrix} b_{11} & b_{12} \\ b_{21} & b_{22} \end{matrix} \right]=\left[ \begin{matrix} a_{11} b_{11}+a_{12} b_{21} & a_{11} b_{12}+a_{12} b_{22} \\ a_{21} b_{11}+a_{22} b_{21} & a_{21} b_{12}+a_{22} b_{22} \end{matrix} \right]$$

Exchanging $a$ and $b$ we obtain the expression for the other product $BA$.

Now, if we allow zero, negative and/or rational entries we can probably decompose any matrix in an infinite number of ways.

However, if we restrict ourselves:

$$a_{jk},~b_{jk} \in \mathbb{N}$$

The problem becomes well defined.

Is there an algorithm to decompose an arbitrary square positive integer matrix into a product of several positive integer matrices of the same dimensions?

There is a set of matrices which can'be be decomposed, just like the prime numbers (or irreducible polynomials, for example). The most trivial one is (remember, zero entries are not allowed):

$$\left[ \begin{matrix} 1 & 1 \\ 1 & 1 \end{matrix} \right]$$

There are no natural numbers $a_{11},b_{11},a_{12},b_{21}$, such that:

$$a_{11} b_{11}+a_{12} b_{21}=1$$

The same extends to any dimension $d$. Any 'composite' $d \times d$ matrix will have all entries $ \geq d$. Thus, for square matrices we can name several more 'primes':

$$\left[ \begin{matrix} 2 & 1 \\ 1 & 1 \end{matrix} \right],~~~\left[ \begin{matrix} 1 & 2 \\ 1 & 1 \end{matrix} \right],~~~\left[ \begin{matrix} 1 & 1 \\ 2 & 1 \end{matrix} \right],~~~\left[ \begin{matrix} 1 & 1 \\ 1 & 2 \end{matrix} \right],~~~\left[ \begin{matrix} 2 & 2 \\ 1 & 1 \end{matrix} \right],~~~\left[ \begin{matrix} 1 & 1 \\ 2 & 2 \end{matrix} \right], \dots$$

And in general, any matrix which has at least one entry equal to $1$.

It makes sense, that most entries in 'composite' matrices will be large, since we are multiplying and adding natural numbers. For example:

$$\left[ \begin{matrix} 1 & 2 & 4 \\ 3 & 3 & 1 \\ 3 & 4 & 4 \end{matrix} \right] \left[ \begin{matrix} 2 & 5 & 5 \\ 4 & 5 & 5 \\ 5 & 1 & 4 \end{matrix} \right]=\left[ \begin{matrix} 30 & 19 & 31 \\ 23 & 31 & 34 \\ 42 & 39 & 51 \end{matrix} \right]$$

$$\left[ \begin{matrix} 2 & 5 & 5 \\ 4 & 5 & 5 \\ 5 & 1 & 4 \end{matrix} \right] \left[ \begin{matrix} 1 & 2 & 4 \\ 3 & 3 & 1 \\ 3 & 4 & 4 \end{matrix} \right] =\left[ \begin{matrix} 32 & 39 & 33 \\ 34 & 43 & 41 \\ 20 & 29 & 37 \end{matrix} \right]$$

If no decomposition algorithm for this case exists, is it at least possible to recognize a matrix that can't be decomposed according to the above rules?

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  • $\begingroup$ Given an integer positive square matrix $C$, we want to find integer positive square matrices $A$ and $B$ such that $C = AB$. In the $2 \times 2$ case, this yields $4$ bilinear Diophantine equations in $4+4$ unknowns. $\endgroup$ – Rodrigo de Azevedo Jul 19 '16 at 16:29
  • $\begingroup$ @RodrigodeAzevedo, thanks. Looks tough, however, for some particular cases solvable. I'm more interested in finding 'primes' with entries $ \geq 2$. I'll post some results in an answer if I succeed $\endgroup$ – Yuriy S Jul 19 '16 at 17:54
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It's a strange question... Let $A\in M(N)$ s.t. $A=PQ$ where $P,Q\in M(N)$ are random. I calculate "the" Smith normal decomposition of $A$: $A=UDV$ where $U,V\in GL(\mathbb{Z})$ and $D$ is a diagonal in $M(\mathbb{Z})$. During each Maple test, I consider the matrix $UD=[C_1,\cdots,C_n]$, where $(C_i)_i$ are its columns; curiously,

(P) for every $i$, $C_i\geq 0$ or $C_i\geq 0$. Is it true for any such matrices $A$ ?

EDIT. Answer to @ You're In My Eye . I conjectured that property (P) above and, for every $i,j$, $a_{i,j}\geq n$ characterize the decomposable matrices $A\in M(N)$. Unfortunately, the matrix $A=\begin{pmatrix}10&13\\9&5\end{pmatrix}\in M(N)$ satisfies (P) but is indecomposable.

Remark 1. If $A=UV$ is decomposable, then there are many other decompositions: $A=(UP)(P^TV)$ where $P$ is any permutation.

Remark 2. We can consider the permanent function; if $A=UV$, then $per(A)> per(U)per(V)$ and in particular $per(U)<\dfrac{per(A)}{n!}$. If we look for an eventual decomposition of the $A$ above, then we obtain $\det(U)\in\{\pm 67,\pm 1\}$ and $per(U)\leq 83$.

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  • $\begingroup$ I would be grateful if you explained a little more how your answer relates to the OP $\endgroup$ – Yuriy S Jul 20 '16 at 21:21
  • $\begingroup$ "There are many other decompostions" - but not an infinite number, right? $\endgroup$ – Yuriy S Jul 21 '16 at 11:54
  • $\begingroup$ Right. If $m=\sup((a_{ij}))$, then, for every decomposition $A=UV$, $\sup((u_{ij}))\leq m-n+1$. $\endgroup$ – loup blanc Jul 21 '16 at 12:21
  • $\begingroup$ As I understand, I need to look up Smith normal form. I'll do that, thank you. So far I posted my attempt at the most simple case, which is still very long $\endgroup$ – Yuriy S Jul 21 '16 at 12:27
  • $\begingroup$ @ You're In My Eye , look at this discussion about NMF (decomposition of a non negative integer matrix in product of two non-negative integer matrices): www4.ncsu.edu/~mtchu/Research/Papers/bindec05f.pdf $\endgroup$ – loup blanc Jul 22 '16 at 11:51
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I now understand that an algorithm for the general case is unlikely. Searching the web, I only found several articles, concerned with positive integer matrix decomposition into binary matrices.


I tried to consider a particular case, which seems to be the most simple:

$$C=\left[ \begin{matrix} 2 & c_b \\ c_a & c \end{matrix} \right]$$

Where $c_a,c_b,c \in \mathbb{N}$ - arbitrary natural numbers. This matrix obviously is either 'prime' or can be decomposed into two 'prime' matrices.

$$\left[ \begin{matrix} 2 & c_b \\ c_a & c \end{matrix} \right]=\left[ \begin{matrix} 1 & 1 \\ a_{1} & a_{2} \end{matrix} \right] \left[ \begin{matrix} 1 & b_{1} \\ 1 & b_{2} \end{matrix} \right]$$

$$ \begin{cases} a_1+a_2=c_a \\ b_1+b_2=c_b \\ a_1b_1+a_2b_2=c \end{cases} $$

Again $a_1,b_1,a_2,b_2 \in \mathbb{N}$. From which we immediately obtain the most important condition for $C$ to be 'composite':

$$ \max(c_a,c_b)\leq c<c_ac_b$$

The equality on the left is possible only in the trivial cases of $a_1=a_2=1$ or $b_1=b_2=1$, which I'm not going to consider.

If this conditions doesn't hold, then $C$ is 'prime'. Which gives us new examples of 'prime' matrices:

$$\left[ \begin{matrix} 2 & 5 \\ 7 & 39 \end{matrix} \right],~~~\left[ \begin{matrix} 2 & 15 \\ 8 & 11 \end{matrix} \right],~~~\dots$$


Another property is this - if we simultaneously permute $a_k$ and $b_k$, then $C$ doesn't change. Which means, that if $C$ is composite, it has at least two factorizations:

$$\left[ \begin{matrix} 2 & c_b \\ c_a & c \end{matrix} \right]=\left[ \begin{matrix} 1 & 1 \\ a_{1} & a_{2} \end{matrix} \right] \left[ \begin{matrix} 1 & b_{1} \\ 1 & b_{2} \end{matrix} \right]=\left[ \begin{matrix} 1 & 1 \\ a_{2} & a_{1} \end{matrix} \right] \left[ \begin{matrix} 1 & b_{2} \\ 1 & b_{1} \end{matrix} \right]$$

Thus, without loss of generality, we can impose:

$$a_1 >a_2,~~~~\frac{c_a}{2}<a_1<c_a \tag{*}$$

The trivial case $a_1=a_2$ gives nilpotent matrix $C$, which I'm not going to consider here.

Solving the system of equations above for $b_1$, we obtain:

$$b_1=\frac{c-c_b(c_a-a_1)}{2a_1-c_a} \tag{**}$$

$$c_a-\frac{c}{c_b}<a_1<c_a$$

The above means that both the intervals $(\frac{c_a}{2},c_a)$ and $(c_a-\frac{c}{c_b},c_a)$ should contain at least one integer:

$$\frac{c_a}{2}>1,~~~\frac{c}{c_b}\geq 2$$

With the condition $(*)$ we have two distinct cases:

$$b_1>b_2,~~~\frac{c_b}{2}<b_1<c_b,~~~~c_b-\frac{c}{c_a}<b_1 \tag{1}$$

$$c>\frac{c_ac_b}{2},~~~~~\frac{c}{c_a} \geq 2,~~~~~\frac{c_b}{2}>1$$

$$b_1<b_2,~~~b_1<\frac{c_b}{2},~~~~b_1<c_b-\frac{c}{c_a} \tag{2}$$

$$c<\frac{c_ac_b}{2},~~~~~\frac{c_b}{2}>1$$

The conditions on $c$ follow from the rearrangement inequality.

There seems to be no additional condition for the case $(2)$, but I'm not sure.

As an example of a composite matrix:

$$C=\left[ \begin{matrix} 2 & 5 \\ 7 & 19 \end{matrix} \right]$$

This is case $(1)$ and all the necessary conditions hold.

$$\frac{c_a}{2}=3.5,~~~~c_a-\frac{c}{c_b}=3.2 \rightarrow a_1 = 4,5,6$$

$$\frac{c_b}{2}=2.5,~~~~c_b-\frac{c}{c_a}=2\frac{4}{7} \rightarrow b_1 = 3,4$$

Using $(**)$ we obtain:

$$a_1=4 \to b_1=4$$

$$a_1=5 \to b_1=3$$

Which gives us (I think) all the solutions:

$$\left[ \begin{matrix} 2 & 5 \\ 7 & 19 \end{matrix} \right]=\left[ \begin{matrix} 1 & 1 \\ 4 & 3 \end{matrix} \right] \left[ \begin{matrix} 1 & 4 \\ 1 & 1 \end{matrix} \right]=\left[ \begin{matrix} 1 & 1 \\ 3 & 4 \end{matrix} \right] \left[ \begin{matrix} 1 & 1 \\ 1 & 4 \end{matrix} \right]$$

$$\left[ \begin{matrix} 2 & 5 \\ 7 & 19 \end{matrix} \right]=\left[ \begin{matrix} 1 & 1 \\ 5 & 2 \end{matrix} \right] \left[ \begin{matrix} 1 & 3 \\ 2 & 1 \end{matrix} \right]=\left[ \begin{matrix} 1 & 1 \\ 2 & 5 \end{matrix} \right] \left[ \begin{matrix} 1 & 2 \\ 1 & 3 \end{matrix} \right]$$


This case was considered to illustrate how I would approach this problem. Which is very long and complicated way. If someone knows better and faster ways, I would be grateful.

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