Show that the direct sum of a kernel of a projection and its image create the originating vector space.

Solution 1:

Take $x \in V$. Since $P=P^2$ we must have $Px=P^2x$ and so $P(x-Px)=0$. Hence $x-Px=\xi$ for some $\xi \in \operatorname{Ker}P$. Thus $x = Px + \xi$. This shows that $V=\operatorname{Im}P + \operatorname{Ker}P$. Now take $y \in \operatorname{Im}P \cap \operatorname{Ker}P$. Since $y \in \operatorname{Im}P$ we have $y=Pz$ for some $z \in V$. Applying $P$ to both sides we get $Py=P^2z$. But $y \in \operatorname{Ker}P$, hence $0=Py=P^2z=Pz=y$. This shows that $\operatorname{Im}P \cap \operatorname{Ker}P=\{0\}$ and so we have $V=\operatorname{Im}P \oplus \operatorname{Ker}P$.

Solution 2:

Hint: $V = \operatorname{Ker}P \oplus \operatorname{Im}P$ iff every $v\in V$ has a unique representation as $v = u+w$ for some $u \in \operatorname{Ker}P, w \in \operatorname{Im}P$ (If you haven't seen that already, it's not too hard to prove.)

How can you find such an expression for general $v$?