Linear Transformation as Matrix Product

Theorem
Let $T: \R^n \to \R^m, \mathbf x \mapsto T\left({\mathbf x}\right)$ be a linear transformation.

Then $T\left({\mathbf x}\right) = \mathbf A_T \mathbf x$, where $\mathbf A_T$ is the $m \times n$ matrix:

where $\left \{ \mathbf {e_i}: 1 \le i \le n \right \}$ is the standard basis of $\R^n$.

Proof
Let $\mathbf x = \begin{bmatrix} x_1 \\ x_2 \\ \vdots \\ x_n \end{bmatrix}$.

Let $\mathbf {I_n}$ be the identity matrix of order $n$.

That $\mathbf A_T$ is $m \times n$ follows from each $T\left({\mathbf {e_i}}\right)$ being a member of $\R^m$ and thus having $m$ rows.

Also see

 * Matrix Product as Linear Transformation