Problem
Show that $\lambda (x + y) = \lambda x + \lambda y$ for all $\lambda \in \mathbb{F}$ and all $x, y \in \mathbb{F}^n$
Solution
Multiplication of an element in $\mathbb{F}^n$ by a scalar is defined coordinate-wise and by problem 9 we know that multiplication in $\mathbb{F}$ is distributive, therefore: $$\lambda (x + y) \stackrel{D1.10 \land D1.12}{=} \lambda (x_1 + y_1, x_2 + y_2, \dots, x_n + y_n)$$ $$\stackrel{D1.17}{=} (\lambda (x_1 + y_1), \lambda (x_2 + y_2), \dots, \lambda (x_n + y_n))$$ $$\stackrel{P1A9}{=} (\lambda x_1 + \lambda y_1, \lambda x_2 + \lambda y_2, \dots, \lambda x_n + \lambda y_n)$$ $$\stackrel{D1.12}{=} (\lambda x_1, \lambda x_2, \dots, \lambda x_n) + (\lambda y_1, \lambda y_2, \dots, \lambda y_n) \stackrel{D1.10 \land D1.12}{=} \lambda x + \lambda y$$ Proving the distributive property of multiplying elements in $\mathbb{F}^n$ by scalars.