Linear Transformation as Matrix Product

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Theorem

Let $T: \R^n \to \R^m, \mathbf x \mapsto \map T {\mathbf x}$ be a linear transformation.

Then:

$\map T {\mathbf x} = \mathbf A_T \mathbf x$

where $\mathbf A_T$ is the $m \times n$ matrix defined as:

$\mathbf A_T = \begin {bmatrix} \map T {\mathbf e_1} & \map T {\mathbf e_2} & \cdots & \map T {\mathbf e_n} \end {bmatrix}$

where $\tuple {\mathbf e_1, \mathbf e_2, \cdots, \mathbf e_n}$ is the standard ordered 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 unit matrix of order $n$.

Then:

\(\ds \mathbf x_{n \times 1}\) \(=\) \(\ds \mathbf I_n \mathbf x_{n \times 1}\) Definition of Left Identity
\(\ds \) \(=\) \(\ds \begin {bmatrix} 1 & 0 & \cdots & 0 \\ 0 & 1 & \cdots & 0 \\ \vdots & \vdots & \ddots & \vdots \\ 0 & 0 & \cdots & 1 \\ \end{bmatrix}\begin{bmatrix} x_1 \\ x_2 \\ \vdots \\ x_n \end {bmatrix}\) Unit Matrix is Identity:Lemma
\(\ds \) \(=\) \(\ds \begin {bmatrix} \mathbf e_1 & \mathbf e_2 & \cdots & \mathbf e_n \end {bmatrix} \begin {bmatrix} x_1 \\ x_2 \\ \vdots \\ x_n \end {bmatrix}\) Definition of Standard Ordered Basis on Vector Space
\(\ds \) \(=\) \(\ds \sum_{i \mathop = 1}^n \mathbf e_i x_i\) Definition of Matrix Product (Conventional)
\(\ds \map T {\mathbf x}\) \(=\) \(\ds \map T {\sum_{i \mathop =1}^n \mathbf e_i x_i}\)
\(\ds \) \(=\) \(\ds \sum_{i \mathop = 1}^n \map T {\mathbf e_i} x_i\) Definition of Linear Transformation on Vector Space
\(\ds \) \(=\) \(\ds \begin {bmatrix} \map T {\mathbf e_1} & \map T {\mathbf e_2} & \cdots & \map T {\mathbf e_n} \end {bmatrix} \begin {bmatrix} x_1 \\ x_2 \\ \vdots \\ x_n \end {bmatrix}\) Definition of Matrix Product (Conventional)

That $\mathbf A_T$ is $m \times n$ follows from each $\map T {\mathbf e_i}$ being an element of $\R^m$ and thus having $m$ rows.

$\blacksquare$


Also see


Sources

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