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This is essentially the process of verifying that the two posts [Linear Algebra] §Matrices and [Linear Algebra] §The Fundamental Theorem of Linear Algebra from [Linear Algebra] also hold for general \(A\)-modules. Most definitions and theorems are not difficult to state in this general setting, but several complications—such as the possibility that \(A\) may be noncommutative and that we will also treat infinite-dimensional free modules—can make things somewhat more complex.
In this post we define matrices and the operations between them.
Basic Definitions
As in [Linear Algebra], an \(m\times n\) matrix over a ring \(A\) is an arrangement of \(mn\) elements of \(A\) in an \(m\times n\) rectangular array
\[X=\begin{pmatrix}x_{11}&x_{12}&\cdots&x_{1n}\\x_{21}&x_{22}&\cdots&x_{2n}\\\vdots&\vdots&\ddots&\vdots\\x_{m1}&x_{m2}&\cdots&x_{mn}\end{pmatrix}\tag{1}\]We denote the set of all such matrices by \(\Mat_{m\times n}(A)\). More generally, we define as follows.
Definition 1 For arbitrary sets \(I, J\), an \(I\times J\) matrix is a family \((x_{ij})_{(i,j)\in I\times J}\) of elements of \(A\) indexed by \(I\times J\), and we denote the collection of all such matrices by \(\Mat_{I\times J}(A)\).
The difference from the usual definition is that \(I\) and \(J\) may be infinite sets. However, in order to define matrix multiplication and to examine the relationship between matrices and linear maps, these sets must be finite, so the above definition is not a major generalization.
Definition 2 For any \(I\times J\) matrix \(X\) and subsets \(I_0\subseteq I\), \(J_0\subseteq J\), the \(I_0\times J_0\) submatrix of \(X\) is defined as the matrix \((x_{ij})_{(i,j)\in I_0\times J_0}\) consisting of those \(x_{ij}\) with \(i\in I_0\) and \(j\in J_0\). In particular, the \(\{i\}\times J\) submatrix of \(X\) is called the \(i\)th row vector of \(X\), and the \(I\times \{j\}\) submatrix is called the \(j\)th column vector of \(X\).
We also define the following.
Definition 3 For any \(I\times J\) matrix \(X=(x_{ij})_{(i,j)\in I\times J}\), the transpose \(X^t=(x^t_{ji})_{(j,i)\in J\times I}\) of \(X\) is the matrix given by the formula
\[x_{ji}^t=x_{ij}\qquad\text{for all $(i,j)\in I\times J$}\]In particular, for an \(m\times n\) matrix \(X\) as in (1), its transpose is the \(n\times m\) matrix \(X^t\in\Mat_{n\times m}(A)\) given by the formula
\[X^t=\begin{pmatrix}x_{11}&x_{21}&\cdots&x_{m1}\\ x_{12}&x_{22}&\cdots&x_{m2}\\\vdots&\vdots&\ddots&\vdots\\ x_{1n}&x_{2n}&\cdots&x_{mn}\end{pmatrix}\]The following proposition is immediate from the definition.
Proposition 4 \((X^t)^t=X\).
Moreover, the \(i\)th row vector of \(X\) is the \(i\)th column vector of \(X^t\), and the \(j\)th column vector of \(X\) is the \(j\)th row vector of \(X^t\).
Operations on Matrices
The goal of this section is to show that \(\Mat_{I\times J}(A)\) has the structure of an \(A\)-module.
First, we show that \(\Mat_{I\times J}(A)\) has the structure of an abelian group. For two matrices
\[X=(x_{ij})_{(i,j)\in I\times J},\qquad Y=(y_{ij})_{(i,j)\in I\times J}\]in \(\Mat_{I\times J}(A)\), we define their sum by
\[X+Y=(x_{ij}+y_{ij})_{(i,j)\in I\times J}\]Then, by the associativity and commutativity of addition in \(A\), we know that the above formula defines an abelian group structure on \(\Mat_{I\times J}(A)\). The identity element with respect to this addition is the \(I\times J\) matrix all of whose entries are \(0\in A\), and the inverse of \(X\) is
\[-X=(-x_{ij})_{(i,j)\in I\times J}\]Then we can verify the formula
\[(X+Y)^t=X^t+Y^t\]Now we must endow \(\Mat_{I\times J}(A)\) with a left \(A\)-action structure. For any \(\alpha\in A\) and \(X\in\Mat_{I\times J}(A)\) as above, we define this by
\[\alpha X=(\alpha x_{ij})_{(i,j)\in I\times J}\]and the properties required for a left \(A\)-module follow from the fact that \(A\) is a left module over itself. Similarly, we define a right \(A\)-action on \(\Mat_{I\times J}(A)\) by
\[X\alpha=(x_{ij}\alpha)_{(i,j)\in I\times J}\]We can verify that \((\alpha X)^t=\alpha X^t\) and \((X\alpha)^t=X^t\alpha\).
Moreover, \(\Mat_{I\times J}(A)\) is a free \(A\)-module, and its basis is given by the matrices
\[E_{ij}=(e_{kl})_{(k,l)\in I\times J},\qquad \text{$e_{kl}=0$ for all $(k,l)$ except for $e_{ij}=1$}\]Strictly speaking, \(E_{ij}\) carries no information about the size of the matrix, so this is a slight abuse of notation, but since there is little risk of confusion in most cases we use it as is. We then see that \((E_{ij})^t=E_{ji}\).
Matrix Multiplication
As in elementary linear algebra, for two matrices \(X\in\Mat_{I\times J}(A)\) and \(Y\in\Mat_{K\times L}(A)\), the product \(XY\) is defined only when \(J=K\), in which case \(XY\) is an \(I\times L\) matrix. Following the entrywise expression for matrix multiplication examined after [Linear Algebra] §Matrices, ⁋Definition 3, the \((i,l)\)-entry of \(XY\) in this general setting is also defined by the formula
\[(XY)_{il}=\sum_{j\in J} X_{ij}Y_{jl}\tag{2}\]However, since the right-hand side of the above formula need not be a finite sum in general, we always assume that \(J=K\) is a finite set when speaking of matrix multiplication.
Now we can verify the following formulas by computing each entry:
\[(XY)Z=X(YZ),\quad X(Y+Z)=XY+XZ,\quad (X+Y)Z=XZ+YZ\]However, unlike in [Linear Algebra] §Matrices, ⁋Proposition 9, there is a cautionary point regarding the relationship between matrix multiplication and transpose: the formula
\[(XY)^t=Y^tX^t\]does not hold in general. This can be seen immediately by writing out both sides entry by entry, and the source of the problem is clear: namely, \(A\) is not commutative. The above formula would make sense if we viewed the two matrices \(X^t\) and \(Y^t\) on the right-hand side as matrices defined over the opposite ring of \(A\), but since in most cases we will consider \(A\) to be commutative, this is not very significant.
References
[Goc] M.S. Gockenbach, Finite-dimensional linear algebra, Discrete Mathematics and its applications, Taylor&Francis, 2011.
[Lee] 이인석, 선형대수와 군, 서울대학교 출판문화원, 2005.
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