The vector space $M_{m \times n}(F) := F^{\,m \times n}$ of $m$ by $n$ matrices over a field, with entrywise operations

Let $F$ be a field (Field) and let $m, n \in \mathbb{N}$. Recall that a natural number is a von Neumann natural (The natural numbers $\mathbb{N}$ (von Neumann)), so $m = \{0, \dots, m-1\}$ and $n = \{0, \dots, n-1\}$ (On $\mathbb{N}$ the order is membership: $m < n \iff m \in n$), and let $m \times n$ be their cartesian product. An $m$ by $n$ matrix over $F$ is an element of the function space $F^{\,m \times n}$ (The vector space $F^{X}$ of all functions $X \to F$ with pointwise operations, and $F^{n}$ as the case $X = n = \{0, 1, \dots, n-1\}$), that is a function $a : m \times n \to F$; we write

$a_{ij} \;:=\; a(i,j) \qquad (i < m,\; j < n)$

for its entries, $i$ being the row index and $j$ the column index. Write

$M_{m \times n}(F) \;:=\; F^{\,m \times n},$

and $M_n(F) := M_{n \times n}(F)$ for the square case.

Since $M_{m \times n}(F)$ is the function space $F^{\,m \times n}$, it is a vector space over $F$ with the pointwise operations of The vector space $F^{X}$ of all functions $X \to F$ with pointwise operations, and $F^{n}$ as the case $X = n = \{0, 1, \dots, n-1\}$, which read entrywise:

$(a + b)_{ij} = a_{ij} + b_{ij}, \qquad (\lambda a)_{ij} = \lambda\, a_{ij} \qquad (i < m,\; j < n),$

and the zero of $M_{m \times n}(F)$ is the matrix all of whose entries are $0_F$. No verification is needed beyond that already carried out in The vector space $F^{X}$ of all functions $X \to F$ with pointwise operations, and $F^{n}$ as the case $X = n = \{0, 1, \dots, n-1\}$ for an arbitrary index set: this is the case $X = m \times n$.

Both indices start at $0$. The rows are indexed by $i \in m = \{0,\dots,m-1\}$ and the columns by $j \in n = \{0,\dots,n-1\}$, so the entries of a $2$ by $3$ matrix are $a_{00}, a_{01}, a_{02}, a_{10}, a_{11}, a_{12}$.

The degenerate shapes. $\mathbb{N}$ contains $0$, so $m = 0$ and $n = 0$ are genuine cases. If $m = 0$ or $n = 0$ then $m \times n = \varnothing$, so $M_{m \times n}(F)$ has exactly one element, the empty function, and is the zero space; there is no matrix of shape $0$ by $n$ other than that one.

• This item introduces the vector-space structure and the notation, and nothing else. There is no matrix product here, no identity matrix, no ring $M_n(F)$, no determinant and no matrix of a linear map. Those belong to a later page, which must add the product to this object rather than introduce a second notion of matrix; the addition and the scalar multiplication used there are the entrywise ones defined above.

• A matrix is a function, so equality of matrices is equality of functions: $a = b$ exactly when $a_{ij} = b_{ij}$ for all $i < m$ and $j < n$. Nothing below relies on a matrix being written as a rectangular array; the array is a way of displaying the function.

• Why the shape is a pair of naturals rather than a pair of finite index sets. Taking the index set to be $m \times n$ with $m$ and $n$ natural numbers keeps the coordinates of a matrix, of a tuple in $F^{n}$ (The vector space $F^{X}$ of all functions $X \to F$ with pointwise operations, and $F^{n}$ as the case $X = n = \{0, 1, \dots, n-1\}$) and the finite sums of The product $g_0 g_1 \cdots g_{n-1}$ of a finite list in a monoid, by recursion, with the empty product ($n = 0$) equal to the identity all indexed by the same kind of object, which is what makes the boundary cases above statable at all.