Exercise 5.B.4

Proof. Suppose $v\in V$. We have $v=(I-P)v+\mathit{Pv}$. Because $P-P^2=0$, it follows that $P(I-P)v=0$. Therefore $(I-P)v\in \mathrm{null}<!--\; FUNCTION\; APPLICATION\; -->P$. Obviously $\mathit{Pv}\in \mathrm{range}<!--\; FUNCTION\; APPLICATION\; -->P$, thus $v\in \mathrm{null}<!--\; FUNCTION\; APPLICATION\; -->P+\mathrm{range}<!--\; FUNCTION\; APPLICATION\; -->P$ and we have $V\subset \mathrm{null}<!--\; FUNCTION\; APPLICATION\; -->P+\mathrm{range}<!--\; FUNCTION\; APPLICATION\; -->P$. The inclusion in the opposite direction is clearly true, therefore $V=\mathrm{null}<!--\; FUNCTION\; APPLICATION\; -->P+\mathrm{range}<!--\; FUNCTION\; APPLICATION\; -->P$.

To see why this is a direct sum, suppose $u\in \mathrm{null}<!--\; FUNCTION\; APPLICATION\; -->P\cap \mathrm{range}<!--\; FUNCTION\; APPLICATION\; -->P$. Because $u\in \mathrm{range}<!--\; FUNCTION\; APPLICATION\; -->P$, there exists $w$ such that $\mathit{Pw}=u$. But $u\in \mathrm{null}<!--\; FUNCTION\; APPLICATION\; -->P$, so we must have

Therefore $\mathrm{null}<!--\; FUNCTION\; APPLICATION\; -->P\cap \mathrm{range}<!--\; FUNCTION\; APPLICATION\; -->P=\{0\}$ and by 1.45 it is direct sum. □