Suppose $P \in \mathscr{L}(V)$ and $P^2 = P$. Prove that $V = \text{null} P \oplus \text{range} P$ where $V$ is a vector space and $\mathscr{L}$ are the set of operators.
If $$V = U_1 + \dots +U_n$$ and $$\text{dim} V = \text{dim} \ U_1 + \dots \ + \text{dim} \ U_n, $$
then $$V = U_1 \oplus \ … \ \oplus U_n.$$
We already know that $$\text{dim} \ V = \text{dimnull} \ P + \text{dimrange} \ P,$$
so only the former remains to be shown. Let $(u_1, \dots, u_n)$ be a basis of $\text{null}\ P.$ Extend this to a basis of $V$. Writing any vector $v$ in $V$ as $v = u + w $, where $u \in \text{null} \ P$, we must show that $w \in \text{range} \ P$.
We have $Pv = Pw$, or $P(Pv – w) = 0$. Thus $(Pv – w) \in \text{null} \ P.$ For some $u' \in \text{null} \ P,$ we have $w = Pv – u',$ but by construction $u' = 0$. Thus $w \in \text{range} \ P$.
I'm not quite sure this is correct.
Best Answer
There is no asumption on finite dimensionality! Let $v \in V$. $v=(v-Pv)+Pv$. $P(v-Pv)=Pv-P^{2}v=\vec{0}$. So $v-Pv \in$ null $P$. Clearly $Pv \in $ Im $P$. Now, let $v \in$ IM $P \cap$ null $P$. Then $Pv=\vec{0}$ and $v=Pu$ for some $u$. So $\vec{0}=Pv=P^{2}u=Pu=v$. The sum is hence direct!