PROBLEM STATEMENT
Let $V$ be a vector space, and let $U, W$ be subspaces of $V$ such that $V = U \oplus W$. Let $P_U$ be the projection on $U$ in the direction of $W$ and $P_W$ the projection on $W$ in the direction of $U$. Prove:
$P_U + P_W = \operatorname{Id}$, $P_U P_W = P_W P_U = 0$
Reciprocally, given $P_1, P_2 : V \rightarrow V$ that verify $P_1 P_2 = P_2 P_1 = 0$, $P_1+P_2 = \operatorname{Id}$, prove that $V = \operatorname{Im}(P_1) \oplus \operatorname{Im}(P_2)$
If possible, generalize to a family of $n$ subspaces whose direct sum gives the vector space.
ATTEMPT AT A SOLUTION
Take $\{v_1, \ldots, v_n\}$ to be a basis for $V$ such that $\{v_1, \ldots, v_k\}$ and $\{v_{k+1}, \ldots, v_n\}$ are bases for $U$ and $W$ respectively. Also, given the direct sum, we know the intersection of the two subspaces will be 0 and therefore any vector belonging to the intersection will be 0.
So $U \ni u = \alpha_1v_1 + \cdots + \alpha_kv_k$ and $W \ni w = \alpha_{k+1}v_{k+1} + \cdots + \alpha_nv_n$.
I can then decompose any vector $v \in V$ into $v = \underbrace{u}_{\in U} + \underbrace{w}_{\in W}$. At this point my notation might get a little weak, so anyone who can help would be appreciated.
Therefore, $$P_U(v) = P_U(u+w) = \underbrace{P_U(u)}_{\in U} + \underbrace{P_U(w)}_{\in U} = P_U(u) + 0 = u$$ Similar reasoning leads to $P_W(v) = w$.
Then, $P_U(P_W(v)) = P_U(w) = 0 = P_W(u) = P_W(P_U(v))$ and $P_U(v) + P_W(v) = u + w = v$ is what occurs to me for the other part of the proof, but I'm not sure if it's quite right, given that $\operatorname{Id}$ is the identity matrix...
I can see where I'm going for the reciprocal statement as well, but I'm having a little more trouble there. Is it as simple as following my reasoning that $P_1(u) = u \in \operatorname{Im}(P_1)$ leads again to $v = u + w$ and thus $V = \operatorname{Im}(P_1) \oplus \operatorname{Im}(P_1) = U \oplus W$ ?
(I took the liberty of assuming the same structure applied to the second half of the first problem with the second set of projections)
If I'm on the right track, does this argument also work to generalize to a case with $n$ subspaces?
Your argument for $\implies$ is correct.
Let $v \in V$. Since $P_1 + P_2 = \operatorname{Id}$, we have $$v = \underbrace{P_1v}_{\in \operatorname{Im}P_1} + \underbrace{P_1v}_{\in \operatorname{Im}P_2} \in \operatorname{Im}P_1 + \operatorname{Im}P_2$$
On the other hand, let $v \in \operatorname{Im}P_1 \cap \operatorname{Im}P_2$. Then $\exists u,w \in V$ such that $v = P_1u = P_2w$. We have
$$P_1v = P_1P_2w = 0$$ $$P_2v = P_2P_1u = 0$$ because $P_1P_2 = P_2P_1 = 0$. Now again $v = P_1v + P_2v = 0+0 = 0$. Therefore $V = \operatorname{Im}P_1 \oplus \operatorname{Im}P_2$.
This generalizes to a family of $n$ operators $P_1, \ldots, P_n : V \to V$ such that $P_1 + \cdots + P_n = \operatorname{Id}$ and $P_iP_j = 0, \forall i\ne j$. You can follow the same argument as above, or you can use induction because from the above assumptions follows $(P_1 + \cdots + P_{n-1}) + P_n = \operatorname{Id}$ and $(P_1 + \cdots + P_{n-1})P_n = P_n(P_1 + \cdots + P_{n-1}) = 0$.
You will get $\operatorname{Im}(P_1+\cdots + P_{n-1}) \oplus \operatorname{Im}(P_n) = V$ and then you can notice that on the space $\operatorname{Im}(P_1+\cdots + P_{n-1}) = \ker P_n$ we have $P_1 + \cdots + P_{n-1} = \operatorname{Id}$ and $P_iP_j = 0, \forall i\ne j$ so induction gives you $\operatorname{Im}(P_1+\cdots + P_{n-1}) = \operatorname{Im}(P_1)\oplus \cdots \oplus \operatorname{Im}(P_{n-1})$.