Can one factor matrices?

Question

I know that one can factor integers as a product of prime numbers. Is there an analog of it to matrices? Can we define prime matrices such that every matrix is a product of prime matrices? Is there any applications of factorization of matrices?

Answer 1

in the ring $D_n (\Bbb{Z})$ of diagonal matrix, we can try to define a prime matrix as follows: $(d_i)$ is prime if $(d_i)$ not null matrix , not unit matrix and all non zero $d_i$ are prime numbers in $\Bbb{Z}$. this type of matrix ring is commutative ring but not integre, and that matrix is decomposed into prime matrix, this decomposition is induced by the decomposition in $\Bbb{Z}$

Answer 2

One can surely "factor" matrices, but the most well-known and useful factorizations doesn't have anything to do with "prime" matrices. I myself have not really come across this concept (that was pointed out in one of the comments). I know that prime ideals are a kind of generalization of prime numbers, but I will leave comments regarding this type of factorization to experts in abstract algebra.

An excellent article on matrix factorization is the following: Bad products of good matrices. written by Paul Halmos.

Factorization is one of the central concerns within research in matrix theory and linear algebra - here are som typical examples:

1. Square-zero factorization of matrices
2. Factorization of singular matrices
3. Idempotent factorization of matrices
4. Nilpotent factorization of matrices
5. Matrix division with an idempotent divisor or quotient

If you search you will find many more examples. The most general question that is asked I suppose, is whether a matrix can be decomposed such that it has a single factor with a certain property (the idempotent divisor article above is concerned with this type of question) - in this case the matrix in question need not be square. But the most popular questions usually center around operators / square matrices and whether they can be decomposed exclusively into factors satisfying some property.

Going further - most of the canonical forms in matrix theory are really just factorizations: diagonalization, Jordan form, rational canonical form, row reduction (row equivalence) - all of these are factorizations that in the end serves to reveal properties such as rank of the relevant matrix, bases for the fundamental spaces, etc. So I think, to summarize, its a resounding "yes": one can factor matrices, in fact it is very central to this field of study in general, but the main concern is not so much to satisfy an exact analogue to factorization of integers into primes. The factors have other properties of interest, i.e. diagonal, triangular, nilpotent, idempotent... again, maybe this can be formally generalized to some equivalent of prime factorization by an expert in abstract algebra.

Answer 3

There are many ways to do so. We can view prime numbers in two different ways to consider their generalizations to other rings, including matrix rings.

The first uses the notion of ideals. Given a commutative ring $R$, an ideal is a subgroup $I$ of $R$ such that $aI\subset I$ for all $a\in R$.

• A prime ideal is an ideal $\mathfrak{p}$ such that for all $a,b\in R$ we have $ab \in R$ if and only if $a\in R$ or $b\in R$.
• An irreducible element is an element $\alpha\in R$ such that for all $a,b\in R$ we have $ab =\alpha$ if and only if $a\in R^ \times$ or $b\in R^\times$.

For the integers $\mathbb{Z}$, an number $p$ is a prime number if and only if $p$ is irreducible if and only if $p\mathbb{Z}$ is a non-zero prime ideal. The fundamental theorem of arithmetic states that every non-zero ideal $I$ can be written uniquely as $$ I=\prod_{\mathfrak{p}_i \text{ prime ideal}}\mathfrak{p_i}^{n_i}.$$

This theorem holds for any Dedekind domain, so we can view factorisation into primes in greater commutative rings as well.

In the ring $\mathbb{Z}$ an integer can also be written uniquely as a product of irreducible elements. In greater rings this need not be the case however. In the ring $R=\mathbb{Z}[\sqrt{-5}]$ (which is a Dedekind domain), we have $$ (1-\sqrt{-5})(1+\sqrt{-5})=6=2\cdot 3$$ with all stated factors being irreducible. In fact in other rings, for example $R=\mathbb{Z}[\sqrt{-17}]$, there exist numbers that can be written as product of both 2 and 3 irreducible elements. (Theorem 5.16)

This notion of irreducible elements also applies to the matrix rings. Provided we consider matrix rings over a ring and not a field. For example, in the ring $R=M_2(\mathbb{Z})$, we have $$\begin{pmatrix} 2 & 0\\ 0 & 2\end{pmatrix}=\begin{pmatrix} 2 & 0\\ 0 & 1\end{pmatrix}\cdot \begin{pmatrix} 1 & 0\\ 0 & 2\end{pmatrix},$$

with the two stated matrices being irreducible. Since $\mathbb{Z}$ is a principal ideal domain, in fact any other irreducible factor will be of the form $U_1AU_2$ where $A$ is any of the matrices above, and $U$ is an invertible matrix (i.e. a matrix of determinant $\pm 1$). (Theorem 5.19) In other rings this is not necessarily the case. Let $K$ be a field and consider the ring $M_2(K[X,Y,Z])$. Then the matrix $$A=\begin{pmatrix}X^2 & XY+Z\\ XY-Z & Y^2\end{pmatrix}$$ is irreducible of determinant $Z^2$ (Example 5.21), hence so is its adjugate matrix. Thus, we have $$A\text{Adj(A)}=\begin{pmatrix} Z^2 & 0\\ 0 & Z^2\end{pmatrix}=\begin{pmatrix} Z & 0\\ 0 & 1\end{pmatrix}^2\cdot \begin{pmatrix} 1 & 0\\ 0 & Z\end{pmatrix}^2,$$ with the matrices on the left and right side of the equation being irreducible.

The theory of factorisations of ideals into prime ideals in commutative rings is a well studied topic in Number Theory, and can be found in almost any book covering that topic. Factorisation theory in non-commutative rings is not as well studied, but a summary can be found here.