Proving any linear transformation can be represented as a matrix

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.

• It's the correct proof- very well written. Sep 1, 2014 at 18:27
• The key here is that $e_1,...,e_n$ is a basis for $\mathbb{R}^n$. Sep 1, 2014 at 19:00
• @copper.hat ah—$\mathbf e_i$ must form a basis so that $\mathbf x$ can be constructed as a linear combination thereof, right? Should I prove that they form a basis? At the risk of jumping to conclusions…isn't it pretty obvious from the definitions of vector addition and scalar multiplication? Sep 1, 2014 at 19:16
• I think it is fairly clear that the $e_k$ form a basis. Sep 1, 2014 at 19:21
• The crux of the proof is that using the standard basis and by linearity, $\mathbf x=\sum\mathbf e_ix_i\implies T(\mathbf x)=\sum T(\mathbf e_i)x_i=\sum\mathbf a_ix_i$ where the $\mathbf a_i$ can be arranged as the columns of the matrix.
– user65203
May 17, 2016 at 19:13