TITLE: Proving any linear transformation can be represented as a matrix
QUESTION [24 upvotes]: I'm trying to prove that

Theorem. Consider a linear transformation $T : \mathbb R^n \to \mathbb R^n$.
  The transformation $T$ can be represented as a matrix product $\mathbf x \mapsto A \mathbf x$, for some matrix $A \in \mathbb R^{n \times n}$.

Here's my attempt at a constructive proof. 

Proof. Consider a matrix $\mathbf x \in \mathbb R^n$ given by
  \begin{align*}
  \mathbf x &=
  \begin{bmatrix}
    x_1 \\
    x_2 \\
    \vdots \\
    x_n
  \end{bmatrix}.
\end{align*}
  We will construct a matrix $A \in \mathbb R^{n \times n}$ such that $T(\mathbf x) = A \mathbf x$.
The vector $\mathbf x$ can also be written as
  \begin{align*}
  \mathbf x &=
  x_1
  \begin{bmatrix}
    1 \\
    0 \\
    \vdots \\
    0
  \end{bmatrix}
  +
  x_2
  \begin{bmatrix}
    0 \\
    1 \\
    \vdots \\
    0
  \end{bmatrix}
  + \dotsb +
  x_n
  \begin{bmatrix}
    0 \\
    0 \\
    \vdots \\
    1
  \end{bmatrix} \\
  &= x_1 \mathbf{e}_{1} + x_2 \mathbf{e}_{2} + \dotsb + x_n \mathbf{e}_{n} \\
  &= \sum_{i=1}^{n} x_i \mathbf{e}_{i},
\end{align*}
  where $\mathbf{e}_{i}$ are the standard basis vectors in $\mathbb R^n$.
Consider the transformation $T(\mathbf x)$.
  Rewriting $\mathbf x$ as above, we have
  \begin{align}
  T(\mathbf x) &= T \left( \sum_{i=1}^{n} x_i \mathbf{e}_{i} \right) \\
  &= \sum_{i=1}^{n} T(x_i \mathbf{e}_{i}) \\
  T(\mathbf x) &= \sum_{i=1}^{n} x_i T(\mathbf{e}_{i}). \tag{1}
\end{align}
Let the matrix $A \in \mathbb R^{n \times n}$ be defined by
  \begin{align*}
  A &=
  \begin{bmatrix}
    T(\mathbf{e}_{1}) &
    T(\mathbf{e}_{2}) &
    \cdots &
    T(\mathbf{e}_{n}) &
  \end{bmatrix} \\
  &=
  \begin{bmatrix}
    a_{11} & \cdots & a_{1n} \\
    \vdots & \ddots & \vdots \\
    a_{n1} & \cdots & a_{nn}
  \end{bmatrix},
\end{align*}
  where each $T(\mathbf{e}_{i})$ is a column of $A$, and each $a_{ij} = T(\mathbf{e}_{i}) \cdot \mathbf{e}_{j}$ is the $j$th component of $T(\mathbf{e}_{i})$.
  Then, by the definition of matrix-vector multiplication, we have
  \begin{align*}
  A \mathbf x &=
  \begin{bmatrix}
    a_{11} & \cdots & a_{1n} \\
    \vdots & \ddots & \vdots \\
    a_{n1} & \cdots & a_{nn}
  \end{bmatrix}
  \begin{bmatrix}
    x_1 \\
    \vdots \\
    x_n
  \end{bmatrix} \\
  &=
  \begin{bmatrix}
    x_1 a_{11} + \dotsb + x_n a_{1n} \\
    \vdots \\
    x_1 a_{n1} + \dotsb + x_n a_{nn} \\
  \end{bmatrix} \\
  &=
  x_1
  \begin{bmatrix}
    a_{11} \\
    \vdots \\
    a_{n1}
  \end{bmatrix}
  + \dotsb +
  x_n
  \begin{bmatrix}
    a_{n1} \\
    \vdots \\
    a_{nn}
  \end{bmatrix} \\
  &= x_1 T(\mathbf{e}_{1}) + \dotsb + x_n T(\mathbf{e}_{n}) \\
  A \mathbf x &= \sum_{i=1}^{n} x_i T(\mathbf{e}_{i}). \tag{2}
\end{align*}
Therefore, by eqs. (1) and (2), we have that
  \begin{align*}
  T(\mathbf x) &= \sum_{i=1}^{n} x_i T(\mathbf{e}_{i}) &
  A \mathbf x &= \sum_{i=1}^{n} x_i T(\mathbf{e}_{i}),
\end{align*}
  and we reach $T(\mathbf x) = A \mathbf x$, as was to be shown.

Any thoughts or suggestions would be appreciated.

REPLY [3 votes]: The proof is correct as written.