Exercise 5.5

Let \(\boldsymbol{a}_1, \boldsymbol{a}_2 \in \mathbb{R}^2\), and define

\[\begin{split}f = \boldsymbol{x} \mapsto \begin{bmatrix}\boldsymbol{a_1} \cdot \boldsymbol{x} \\ \boldsymbol{a_2} \cdot \boldsymbol{x} \end{bmatrix}\end{split}\]

Then,

\[\begin{split}f(\boldsymbol{x} + \boldsymbol{y}) &= \begin{bmatrix} \boldsymbol{a_1} \cdot (\boldsymbol{x} + \boldsymbol{y}) \\ \boldsymbol{a_2} \cdot (\boldsymbol{x} + \boldsymbol{y}) \end{bmatrix} \\ &= \begin{bmatrix} \boldsymbol{a_1} \cdot \boldsymbol{x} + \boldsymbol{a_1} \cdot \boldsymbol{y} \\ \boldsymbol{a_2} \cdot \boldsymbol{x} + \boldsymbol{a_2} \cdot \boldsymbol{y} \\ \end{bmatrix} \\ &= \begin{bmatrix} \boldsymbol{a_1} \cdot \boldsymbol{x} \\ \boldsymbol{a_2} \cdot \boldsymbol{x} \end{bmatrix} + \begin{bmatrix} \boldsymbol{a_1} \cdot \boldsymbol{y} \\ \boldsymbol{a_2} \cdot \boldsymbol{y} \end{bmatrix} \\ &= f(\boldsymbol{x}) + f(\boldsymbol{y})\end{split}\]

The important thing to note about this proof is that we are using the property which we previously proved about the dot product, that \(\boldsymbol{x} \cdot (\boldsymbol{y} + \boldsymbol{z}) = \boldsymbol{x} \cdot \boldsymbol{y} + \boldsymbol{x} \cdot \boldsymbol{z}\).

Similarly,

\[\begin{split}f(k \boldsymbol{x}) &= \begin{bmatrix} \boldsymbol{a_1} \cdot (k\boldsymbol{x}) \\ \boldsymbol{a_2} \cdot (k\boldsymbol{x}) \end{bmatrix} \\ &= \begin{bmatrix} k (\boldsymbol{a_1} \cdot \boldsymbol{x}) \\ k (\boldsymbol{a_2} \cdot \boldsymbol{x}) \end{bmatrix} \\ &= k \begin{bmatrix} \boldsymbol{a_1} \cdot \boldsymbol{x} \\ \boldsymbol{a_2} \cdot \boldsymbol{x} \end{bmatrix} \\ &= k f(\boldsymbol{x})\end{split}\]

This argument similarly uses the other property of the dot product which we proved a moment ago, namely that \((k_1 \boldsymbol{x}) \cdot (k_2 \boldsymbol{y}) = k_1 k_2 (\boldsymbol{x} \cdot \boldsymbol{y})\).

We have proved that the two linear mapping laws hold for any such function \(f\), and therefore we are done: any function defined in terms of dot products like this is a linear mapping.