Linear Complements and endomorphisms Let $U,W\subseteq V$ be linear subspaces of a finite-dimensional vectorspace $V$.
I need to show that the following are equivalent:


*

*$V=U\bigoplus W$

*There exists an endomorphism $\varphi:V\to V$ s.t.: (a) $\varphi(\varphi(v))=\varphi(v)$ f.a. $v\in V$, (b) $image(\varphi)=U$, (c) $ker(\varphi)=W$

*There exist two endomorphisms $\pi_1,\pi_2:V\to V$ s.t.: (a) $image(\pi_1)\subseteq U$, $image(\pi_2)\subseteq W$, (b) $\pi_1+\pi_2=id_V$, (c) $\pi_1\upharpoonright U=id_U$, $\pi_2\upharpoonright W=id_W$


I've gone as far as showing some connection between 1 and 2 via the properties of bases for $U,W$ as $dim(V)=dim(ker(\varphi))+dim(image(\varphi))=dim(W)+dim(U)$ which should imply that they are linear complements through the dimension formula for direct products.
 A: This problem is actually about projection. A projection on a vector space is defined as an endomorphism $P:V \to V$ such that $P^2 = P$. If $P$ is an projection on $V$, then for any $v \in V$,
$$v = Pv + (I-P)v$$
This implies
$$V=image(P)+image(I-P)$$
Moreover, it can be proven that
$$image(I-P)=ker(P)$$
since $P(I-P)v=(P-P^2)v=0 \implies image(I-P) \subset ker(P)$
$$v \in ker(P) \implies v=Pv+(I-P)v=(I-P)v \in image(I-P) \implies ker(P) \subset image(I-P)$$
It remains to show that $image(P) \cap ker(P)= \{0\}$. Let $Pv \in image(P) \cap ker(P)$. Then $P(Pv)=Pv=0$. Thus we have shown
$$V=image(P) \oplus ker(P)$$
Conversely, if $V=U \oplus W$, then for any $v \in V $, $v=u+w$, define $P:V \to V$ by $Pv=u$. It is not hard to verify that $P$ is a projection.
For part 3, simply notice that $\pi_1,\pi_2$ are projections onto $U$ and $W$, respectively. If $P$ is a projection, then $I-P$ is also a projection.
Elaboration of part 3: 
$2) \implies 3)$ Say $\varphi$ is the given projection. Let $\pi_1 = \varphi$, $\pi_2 = id_V-\varphi$. Then simply verify $\pi_1, \pi_2$ satisfies the desired properties.
$3) \implies 1)$ For any $v \in V$, $ v = \pi_1 v + \pi_2 v \in U+
W$. Thus $V = U + W$. Now assume $v \in U \cap W$. By property c), $v \in U$ implies $\pi_1v = v$; $v \in W$ implies $\pi_2 v = v.$ Then
$$v = \pi_1 v + \pi_2 v = v + v \implies v=0$$
Hence $U \cap W = \{ 0 \}$. Therefore $V = U \oplus W$.
$1) \implies 2)$ is already shown above.
