Section 2.3
Matrices
Before initiating the grand project of classifying vector spaces, I would like to talk about matrices, which are generally presented to be the main objects of study in introductory linear algebra textbooks.
A $m \times n$ matrix with entries in $k$ is a $k$-linear map $k^n \to k^m$, where $k^n$ and $k^m$ are regarded as $k$-vector spaces.
What does this mean? A matrix is supposed to be an array with numbers, not a linear map. But let us try to think about how we can encode a $k$-linear map $k^n \to k^m$.
As a warm-up, let us try to classify all $k$-linear maps $T : k \to k$, where both $k$ are considered as $k$-vector spaces. If we demand that $T(1) = c$ for some $c \in k$, then we get
$$ T(a) = T(a \cdot 1) = a \cdot T(1) = a \cdot c = ac $$
for all $a \in k$. That is, the value of $T(1)$ completely determines $T$. On the other hand, once we choose $T(1) = c$ arbitrarily, the map $T : a \mapsto ac$ is indeed a linear map, because
$$ T(a + b) = (a + b) c = a c + bc = T(a) + T(b), \quad T(a \cdot b) = a b c = a (bc) = a T(b). $$
This shows that there is a bijection
$$ \lbrace k\text{-linear maps } k \to k \rbrace \quad \longleftrightarrow \quad k; \qquad T \mapsto T(1), $$
and the linear maps $k \to k$ are completely classified by $k$.
Let $V$ be an arbitrary vector space. Show that the linear maps $T : k \to V$ are classified by elements $V$. That is, exhibit a bijection
$$ \lbrace k\text{-linear maps } k \to V \rbrace \quad \longleftrightarrow \quad V. $$
Let us now try to classify linear maps $k^n \to k^m$. Here, we are going to use the notation
$$ e_1 = (1, 0, \ldots, 0), \quad e_2 = (0, 1, 0, \ldots, 0), \quad \ldots. $$
So, for instance, we can write
$$ (a_1, a_2, \ldots, a_n) = a_1 e_1 + a_2 e_2 + \cdots + a_n e_n \in k^n. $$
If $T : k^n \to k^m$ is a linear map, for each $1 \le j \le n$ we get an element $T(e_j) \in k^m$. Let us write
$$ T(e_j) = (t_{1j}, t_{2j}, \ldots, t_{mj}) = \sum_{i=1}^{m} t_{ij} e_i. $$
From a $k$-linear map $T : k^n \to k^m$, we have thus extracted $mn$ numbers $t_{ij} \in k$ for $1 \le i \le m$ and $1 \le j \le n$. These $mn$ numbers completely determine $T$, because
$$ T((a_1, \ldots, a_n)) = T\biggl( \sum_{j=1}^{n} a_j e_j \biggr) = \sum_{j=1}^{n} a_j T(e_j) = \sum_{j=1}^{n} \sum_{i=1}^{m} a_j t_{ij} e_i. $$
Conversely, for arbitrary numbers $t_{ij} \in k$, the map defined above is $k$-linear, because
$$ \begin{aligned} T((a_1 + b_1, \ldots, a_n + b_n)) &= \sum_{j=1}^{n} \sum_{i=1}^{m} (a_j + b_j) t_{ij} e_i \\ &= \sum_{j=1}^{n} \sum_{i=1}^{m} a_j t_{ij} e_i + \sum_{j=1}^{n} \sum_{i=1}^{m} b_j t_{ij} e_i \\ &= T((a_1, \ldots, a_n)) + T((b_1, \ldots, b_n)) \end{aligned} $$
and
$$ T(a (b_1, \ldots, b_n)) = \sum_{j=1}^{n} \sum_{i=1}^{m} (a b_j) t_{ij} e_i = a \sum_{j=1}^{n} \sum_{i=1}^{m} b_j t_{ij} e_i = a T((b_1, \ldots, b_n)). $$
Hence the linear map $T : k^n \to k^m$ is uniquely determined by the numbers $t_{ij}$ and we get a bijection
$$ \lbrace k\text{-linear maps } T : k^n \to k^m \rbrace \quad \longleftrightarrow \quad \lbrace mn \text{ numbers } t_{ij} \in k \rbrace . \tag{$\ast$} $$
To keep track of all the numbers, we are going to introduce a new notation. Let us write
$$ T = \begin{bmatrix} t_{11} & t_{12} & \cdots & t_{1n} \\ t_{21} & t_{22} & \cdots & t_{2n} \\ \vdots & \vdots & \ddots & \vdots \\ t_{m1} & t_{m2} & \cdots & t_{mn} \end{bmatrix} $$
when a linear map $T : k^n \to k^m$ on the left hand side of ($\ast$) corresponds to $t_{ij} \in k$ on the right hand side. With this notation, we are going to write
$$ T((a_1, \ldots, a_n)) = (b_1, \ldots, b_m) $$
as
$$ \begin{bmatrix} t_{11} & t_{12} & \cdots & t_{1n} \\ t_{21} & t_{22} & \cdots & t_{2n} \\ \vdots & \vdots & \ddots & \vdots \\ t_{m1} & t_{m2} & \cdots & t_{mn} \end{bmatrix} \begin{bmatrix} a_1 \\ a_2 \\ \vdots \\ a_n \end{bmatrix} = \begin{bmatrix} b_1 \\ b_2 \\ \vdots \\ b_m \end{bmatrix}. $$
These are all equivalent to saying that
$$ b_i = \sum_{j=1}^{n} t_{ij} a_j, $$
but with the new notation, we can write out the equations explicitly in a single equation.
The notation seems to suggest that $(a_1, \ldots, a_n)$ is a $1 \times n$ matrix, rather than a vector in $k^n$. In a sense, this is true. Exercise 2.3.A shows that an element of a vector space can be canonically identified with a map from $k$ to the vector space. So an element of $k^n$ can be considered as a $k$-linear map $k \to k^n$, which is an $n \times 1$ matrix. If this is confusing, don't worry about it.
Show that the identity map $\id : k^n \to k^n$ corresponds to the matrix
$$ I = \begin{bmatrix} 1 & 0 & \cdots & 0 \\ 0 & 1 & \cdots & 0 \\ \vdots & \vdots & \ddots & \vdots \\ 0 & 0 & \cdots & 1 \end{bmatrix}. $$
Let $S : k^n \to k^m$ and $T : k^m \to k^p$ be two matrices. ($S$ will be $m \times n$ and $T$ will be $p \times m$.) The composite $T \circ S = U : k^n \to k^p$ is going to be $k$-linear, and hence a matrix. Let $t_{ij}$, $s_{jl}$, $u_{il}$ be the matrix entries corresponding to the linear maps $T$, $S$, $U$. Then show that
$$ u_{il} = \sum_{j=1}^{m} t_{ij} s_{jl}. $$
If you have learned about matrix multiplications before, check that this agrees with the ordinary way of multiplying matrices.
So from the perspective of linear maps, multiplication of matrices is nothing other than composition of linear maps. If you have been bewildered by the fact that matrix multiplication is associative, that is,
$$ A (B C) = (A B) C $$
for all matrices $A$, $B$, and $C$, now you know that this is because composition of maps is associative.