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April 1, 2012

  • Now the Riemann Hypothesis has finally been solved (who would have guessed that it would be soluble relying solely upon analysis of elementary functions?) someone's going to have to go and post it up. I'll get there but I'm bogged down in tidying up the Relation Theory category. --prime mover

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Proof of the Week

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 defined as:

\(\displaystyle \) \(\displaystyle \) \(\displaystyle \) \(\displaystyle \mathbf A_T\) \(=\) \(\displaystyle \begin{bmatrix} \\ \\ \\ T\left({\mathbf e_1 }\right) & T\left({\mathbf e_2 }\right) & \cdots & T\left({\mathbf e_n }\right) \\ \\ \\ \end{bmatrix}\) \(\displaystyle \) \(\displaystyle \) \(\displaystyle \)                    

where $\left \{ {\mathbf e_1, \mathbf e_2, \cdots, \mathbf e_n} \right\}$ 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 $n$.

\(\displaystyle \) \(\displaystyle \) \(\displaystyle \) \(\displaystyle \mathbf x_{n \times 1}\) \(=\) \(\displaystyle \mathbf I_n \mathbf x_{n \times 1}\) \(\displaystyle \) \(\displaystyle \) \(\displaystyle \)          definition of left identity          
\(\displaystyle \) \(\displaystyle \) \(\displaystyle \) \(\displaystyle \) \(=\) \(\displaystyle \begin{bmatrix} 1 & 0 & \cdots & 0 \\ 0 & 1 & \cdots & 0 \\ \vdots & \vdots & \ddots & \vdots \\ 0 & 0 & \cdots & 1 \\ \end{bmatrix}\begin{bmatrix} x_1 \\ x_2 \\ \vdots \\ x_n \end{bmatrix}\) \(\displaystyle \) \(\displaystyle \) \(\displaystyle \)          Identity Matrix is Identity:Lemma          
\(\displaystyle \) \(\displaystyle \) \(\displaystyle \) \(\displaystyle \) \(=\) \(\displaystyle \begin{bmatrix} \\ \\ \\ \mathbf e_1 & \mathbf e_2 & \cdots & \mathbf e_n \\ \\ \\ \end{bmatrix}\begin{bmatrix} x_1 \\ x_2 \\ \vdots \\ x_n \end{bmatrix}\) \(\displaystyle \) \(\displaystyle \) \(\displaystyle \)          definition of standard basis          
\(\displaystyle \) \(\displaystyle \) \(\displaystyle \) \(\displaystyle \) \(=\) \(\displaystyle \sum_{i=1}^n \mathbf e_i x_i\) \(\displaystyle \) \(\displaystyle \) \(\displaystyle \)          definition of matrix multiplication          
\(\displaystyle \) \(\displaystyle \) \(\displaystyle \) \(\displaystyle T\left({\mathbf x}\right)\) \(=\) \(\displaystyle T\left({\sum_{i=1}^n \mathbf e_i x_i}\right)\) \(\displaystyle \) \(\displaystyle \) \(\displaystyle \)                    
\(\displaystyle \) \(\displaystyle \) \(\displaystyle \) \(\displaystyle \) \(=\) \(\displaystyle \sum_{i=1}^n T\left({\mathbf e_i}\right)x_i\) \(\displaystyle \) \(\displaystyle \) \(\displaystyle \)          definition of linear transformation          
\(\displaystyle \) \(\displaystyle \) \(\displaystyle \) \(\displaystyle \) \(=\) \(\displaystyle \begin{bmatrix} \\ \\ \\ T\left({\mathbf e_1}\right) & T\left({\mathbf e_2}\right) & \cdots & T\left({\mathbf e_n}\right) \\ \\ \\ \end{bmatrix} \begin{bmatrix} x_1 \\ x_2 \\ \vdots \\ x_n \end{bmatrix}\) \(\displaystyle \) \(\displaystyle \) \(\displaystyle \)          definition of matrix multiplication          

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.

$\blacksquare$


Also see


Sources

  • For a video presentation of the contents of this page, visit the Khan Academy.


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