A Matrix is a Linear Operator

Take a matrix $A \in F^{m \times n}$ and a vector $x \in F^{n}$ . Multiply them. Each component of the output is a sum:

(A x)_{i} = j = 1 \sum n a_{ij} x_{j} .

Rows prove linearity

Read one row as a fixed recipe. The $i$ -th row of $A$ does not change while the input $x$ changes. It takes the coordinates of $x$ , weights them, and adds the products:

(A x)_{i} = a_{i 1} x_{1} + a_{i 2} x_{2} + \dots + a_{in} x_{n} .

This row-wise formula is enough to prove additivity. For $x, y \in F^{n}$ ,

(A (x + y))_{i} = j = 1 \sum n a_{ij} (x_{j} + y_{j}) = j = 1 \sum n a_{ij} x_{j} + j = 1 \sum n a_{ij} y_{j} = (A x)_{i} + (A y)_{i} .

It also proves homogeneity. For $λ \in F$ ,

(A (λ x))_{i} = j = 1 \sum n a_{ij} (λ x_{j}) = λ j = 1 \sum n a_{ij} x_{j} = λ (A x)_{i} .

Since both identities hold for every output coordinate $i$ , the whole map

x ⟼ A x

is linear. Thus every $m \times n$ matrix defines a linear map $F^{n} \to F^{m}$ .

Columns store the images

Rows explain why the map is linear. Columns explain what the map stores.

Let $c_{j}$ denote the $j$ -th column of $A$ :

A = [c_{1} c_{2} \dots c_{n}] .

Then matrix-vector multiplication can be written as a column combination:

A x = x_{1} c_{1} + x_{2} c_{2} + \dots + x_{n} c_{n} .

The entries of $x$ are not just input coordinates. They are the coefficients that tell how strongly each column of $A$ contributes to the output.

Basis vectors select columns

Let $e_{j}$ be the $j$ -th standard basis vector of $F^{n}$ . Its $k$ -th coordinate is

(e_{j})_{k} = δ_{jk} = {10 k = j, k \neq = j .

Apply $A$ and inspect the $i$ -th coordinate:

(A e_{j})_{i} = k = 1 \sum n a_{ik} (e_{j})_{k} = k = 1 \sum n a_{ik} δ_{jk} = a_{ij} .

Thus

A e_{j} = a_{1 j} a_{2 j} ⋮ a_{mj},

which is exactly the $j$ -th column of $A$ . The $j$ -th column is not merely stored in $A$ ; it is the image of the $j$ -th basis vector.

So the columns of $A$ are the images of the standard basis vectors. Once those images are known, linearity determines the value of the map on every vector.

The reverse direction

A linear map also determines a matrix once a basis has been chosen. Work in the standard basis $e_{1}, \dots, e_{n}$ of $F^{n}$ . Let

T : F^{n} \to F^{m}

be a linear map. Define the columns of a matrix $A$ by

c_{j} := T (e_{j}), A = [c_{1} c_{2} \dots c_{n}] .

Now take an arbitrary vector $x \in F^{n}$ . In the standard basis it decomposes as

x = j = 1 \sum n x_{j} e_{j} .

Because $T$ is linear, it is forced to respect this decomposition:

T (x) = T (j = 1 \sum n x_{j} e_{j}) = j = 1 \sum n x_{j} T (e_{j}) = j = 1 \sum n x_{j} c_{j} = A x .

So a linear map has no freedom beyond its values on the basis vectors. Once the columns $T (e_{1}), \dots, T (e_{n})$ are known, the matrix $A$ reproduces $T$ on every input.

The same object in coordinates

The two constructions are inverse to each other:

start with a matrix $A$ , get the linear map $x \mapsto A x$ ;
start with a linear map $T$ , get the matrix whose columns are $T (e_{1}), \dots, T (e_{n})$ .

Doing both brings you back to where you started. That is the precise sense in which matrices and linear maps between coordinate spaces carry the same information.

Formally, this gives an isomorphism of vector spaces:

Hom (F^{n}, F^{m}) ≅ F^{m \times n} .

The operations match under this correspondence:

adding linear maps adds their matrices column by column;
scaling a linear map scales every column;
composing linear maps becomes matrix multiplication.

There is one important caveat. A matrix is not a basis-free object. It is the coordinate representation of a linear map after choosing bases for the domain and codomain. Change the bases, and the same map may be written by a different matrix.

So a matrix is not merely a table of numbers. Once coordinates are fixed, it is a linear map written down. If the domain and codomain are the same space, it represents a linear operator.

The operator is the action. The matrix is how that action is written after choosing coordinates.

Lukas' Notes