Exercise 5.6

Let \(f, g\) be linear mappings. We consider the function \(f \circ g\), defined as

\[f \circ g = \boldsymbol{x} \mapsto f(g(\boldsymbol{x}).\]

Firstly, we know that

\[f(g(\boldsymbol{x} + \boldsymbol{y})) = f(g(\boldsymbol{x}) + g(\boldsymbol{y}))\]

since \(g\) is a linear mapping by assumption. Now we use the fact that \(f\) is a linear mapping to conclude that

\[f(g(\boldsymbol{x}) + g(\boldsymbol{y})) = f(g(\boldsymbol{x})) + f(g(\boldsymbol{y})).\]

We have therefore shown that \((f \circ g)(\boldsymbol{x} + \boldsymbol{y}) = (f \circ g)(\boldsymbol{x}) + (f \circ g)(\boldsymbol{y})\) and so we have established the first linear mapping law.

The second part of the proof is very similar: we show that \(f \circ g\) is compatible with scalar multiplication by first using the fact that \(g\) is compatible with scalar multiplication and then by using the fact that \(f\) is.