Problem
Suppose $$U = \lbrace (x,y,x+y,x-y,2x) \in \mathbb{F}^5 | x,y \in \mathbb{F} \rbrace$$ Find three subspaces $W_1, W_2, W_3$ of $\mathbb{F}^5$, none of which equals $\lbrace 0 \rbrace$, such that $\mathbb{F}^5 = U \oplus W_1 \oplus W_2 \oplus W_3$.
Solution
We saw in P1C21 that $U \oplus W = \mathbb{F}^5$ if $W = \lbrace (0,0,x,y,z) \in \mathbb{F}^5 | x,y,z \in \mathbb{F} \rbrace$. But clearly $W = W_1 \oplus W_2 \oplus W_3$ if: $$W_1 = \lbrace (0,0,x,0,0) \in \mathbb{F}^5 | x \in \mathbb{F} \rbrace$$ $$W_2 = \lbrace (0,0,0,x,0) \in \mathbb{F}^5 | x \in \mathbb{F} \rbrace$$ $$W_3 = \lbrace (0,0,0,0,x) \in \mathbb{F}^5 | x \in \mathbb{F} \rbrace$$ And by P1C17 plus P1C21 we therefore see that: $$\mathbb{F}^5 \stackrel{P1C21}{=} U \oplus W = U \oplus (W_1 \oplus W_2 \oplus W_3) \stackrel{P1C17}{=} U \oplus W_1 \oplus W_2 \oplus W_3$$