Matrices ↔ Linear Maps

A linear map (or linear transformation) is a function $T:\mathbb{R}^n\to\mathbb{R}^m$ that respects addition and scaling. Here $\mathbb{R}^n$ is the set of vectors (lists) with $n$ real entries — the domain — and $\mathbb{R}^m$ is the codomain. "Respects addition and scaling" means for all vectors $\mathbf{u},\mathbf{v}\in\mathbb{R}^n$ and every scalar (real number) $c$:

$T(\mathbf{u}+\mathbf{v}) = T(\mathbf{u})+T(\mathbf{v}), \qquad T(c\,\mathbf{u}) = c\,T(\mathbf{u}).$

These two rules combine into one: $T(a\mathbf{u}+b\mathbf{v}) = aT(\mathbf{u})+bT(\mathbf{v})$ for all scalars $a,b$.

Here is the key consequence. The standard basis vectors of $\mathbb{R}^n$ are $\mathbf{e}_1=\begin{bmatrix}1\\0\\\vdots\\0\end{bmatrix},\ \mathbf{e}_2=\begin{bmatrix}0\\1\\\vdots\\0\end{bmatrix},\ \dots,\ \mathbf{e}_n=\begin{bmatrix}0\\0\\\vdots\\1\end{bmatrix},$ where $\mathbf{e}_j$ has a $1$ in position $j$ and $0$ everywhere else. Any vector $\mathbf{v}=\begin{bmatrix}v_1\\\vdots\\v_n\end{bmatrix}$ can be written as $\mathbf{v}=v_1\mathbf{e}_1+\cdots+v_n\mathbf{e}_n$. Applying $T$ and using linearity: $T(\mathbf{v}) = v_1\,T(\mathbf{e}_1)+v_2\,T(\mathbf{e}_2)+\cdots+v_n\,T(\mathbf{e}_n).$ So once you know the $n$ output vectors $T(\mathbf{e}_1),\dots,T(\mathbf{e}_n)$, you know $T$ on every input. A linear map is completely determined by what it does to the basis. That is exactly the information a matrix stores.

Collect those output vectors as the columns of a matrix. The standard matrix of a linear map $T:\mathbb{R}^n\to\mathbb{R}^m$ is

$A = \big[\,T(\mathbf{e}_1)\ \ T(\mathbf{e}_2)\ \ \cdots\ \ T(\mathbf{e}_n)\,\big].$

Each $T(\mathbf{e}_j)$ lives in $\mathbb{R}^m$, so it contributes $m$ numbers (one column); there are $n$ of them. Therefore $A$ has $m$ rows and $n$ columns — it is an $m\times n$ matrix. A handy slogan: rows = codomain dimension, columns = domain dimension.

Recipe to build $A$:

1. Plug $\mathbf{e}_1=(1,0,\dots,0)$ into the formula for $T$. The result is column $1$.
2. Plug $\mathbf{e}_2=(0,1,0,\dots,0)$ into $T$. The result is column $2$.
3. Continue through $\mathbf{e}_n$.

Reading one entry. The entry of $A$ in row $i$, column $j$ — written $a_{ij}$ or $A_{ij}$ — is the $i$-th component of the vector $T(\mathbf{e}_j)$. So you can find a single entry without building the whole matrix: evaluate $T(\mathbf{e}_j)$ and look at coordinate $i$.

The matrix $A$ isn't just a storage box — multiplying by it is the map. For every $\mathbf{v}\in\mathbb{R}^n$, $T(\mathbf{v}) = A\mathbf{v}.$

Why? Writing $A=[\,\mathbf{a}_1\ \cdots\ \mathbf{a}_n\,]$ with columns $\mathbf{a}_j=T(\mathbf{e}_j)$, the matrix–vector product is defined as the linear combination of the columns weighted by the entries of $\mathbf{v}$: $A\mathbf{v} = v_1\mathbf{a}_1 + v_2\mathbf{a}_2 + \cdots + v_n\mathbf{a}_n = v_1 T(\mathbf{e}_1)+\cdots+v_n T(\mathbf{e}_n) = T(\mathbf{v}).$ That last equality is exactly the linearity identity from the first section.

In practice you compute $A\mathbf{v}$ row by row: the $i$-th entry of $A\mathbf{v}$ is the dot product of row $i$ of $A$ with $\mathbf{v}$. For example, $\begin{bmatrix}2&3\\1&-1\end{bmatrix}\begin{bmatrix}4\\1\end{bmatrix}=\begin{bmatrix}2\cdot4+3\cdot1\\1\cdot4+(-1)\cdot1\end{bmatrix}=\begin{bmatrix}11\\3\end{bmatrix}.$ For the product $A\mathbf{v}$ to be defined, the number of columns of $A$ must equal the number of entries of $\mathbf{v}$ — which is exactly $n$, the domain dimension. The output has $m$ entries, landing in the codomain $\mathbb{R}^m$.

Suppose you do one linear map and then another. Let $S:\mathbb{R}^n\to\mathbb{R}^p$ have standard matrix $B$ (size $p\times n$), and $T:\mathbb{R}^p\to\mathbb{R}^m$ have standard matrix $A$ (size $m\times p$). The composition $T\circ S:\mathbb{R}^n\to\mathbb{R}^m$ means "apply $S$ first, then $T$": $(T\circ S)(\mathbf{v}) = T\big(S(\mathbf{v})\big)$.

Its standard matrix is the matrix product $AB$: $(T\circ S)(\mathbf{v}) = T(S(\mathbf{v})) = A(B\mathbf{v}) = (AB)\mathbf{v}.$

Note the order carefully: $S$ acts first, but its matrix $B$ sits on the right, because it touches the vector first. The sizes must chain correctly: $A$ is $m\times p$ and $B$ is $p\times n$, so $AB$ is $m\times n$ — the inner dimensions ($p$) match and cancel. This is the deep reason matrix multiplication is defined the way it is: it is engineered so that multiplying matrices mirrors composing the maps they represent. Matrix multiplication is generally not commutative ($AB\neq BA$ in general), which matches the fact that doing $S$ then $T$ usually differs from doing $T$ then $S$.