TITLE: Prove $p^2=p$ and $qp=0$
QUESTION [7 upvotes]: I am not really aware what's going on in this question. I appreciate your help.
Let $U$ be a vector space over a field $F$ and $p, q: U \rightarrow U$ linear maps. Assume $p+q = \text{id}_U$ and $pq=0$. Let $K=\ker(p)$ and $L=\ker(q)$.

Prove $p^2=p$ and $qp=0$.

$$p(p+q) = p(\text{id}_U) \Rightarrow p^2+pq=p \Rightarrow p^2 =p.$$ I actually first found this by letting $(\text{id}_U) =1$, although probably a wrong to do it.
Since $p^2 =p$ and $q$ is also a linear map defined by $q: U \rightarrow U$ then $q^2=q$. Again, I'm not sure if this actually is the right way to do it.
Then we have $$q(p+q) = q(\text{id}_U) \Rightarrow q^2+qp=q \Rightarrow q^2+ qp= q^2 \Rightarrow qp=0.$$

Prove $K=\text{im}(q)$.

For this question, I honestly do not know how to tackle it from the following definitions.
$K = \ker(p)=\left\{u \in U \ |\ p(u)=0_U\right\}$ and $\text{im}(q)=\left\{q(u) \ | \ u \in U \right\}$.
What's the correct way of doing these questions? Thank you for your time.

REPLY [3 votes]: Since $\mbox{id}_U$ is the unit of the ring $L(U)$, it is standard to denote it by $1$.
So you know that $p$ is idempotent ($p^2=p$). It follows automatically that $q=1-p$ is idempotent as $q^2=(1-p)^2=1^2-p-p+p^2=1-2p+p=1-p=q$. And then $qp=(1-p)p=p-p^2=0$ as well. 
Such a pair $(p,q)$ is the $n=2$ case of what is called an orthogonal decomposition of the unit. When such a thing happens, we have the Peirce decomposition which is very handy for many purposes. In operator theory/algebras in particular.
For any given idempotent $p$, the range and the nullspace yield a direct sum decomposition of the vector space $V=\mbox{im} p\oplus \ker p$. Conversely, a direct sum decomposition of $V=I\oplus K$ defines a unique idempotent such that $\mbox{im} p=I$ and $\ker p=K$. Cf. remark below. Now let us answer your question.

The relation between the idempotents $p$ and $q=1-p$ is
  $$
\mbox{im }(1-p)=\ker p\qquad \ker (1-p)=\mbox{im} \,p.
$$

Proof: by symmetry, it suffices to prove one of these two properties. 
Let us do the lhs.
Since $0=p-p^2=p(1-p)$, we get $p(1-p)x=0$ for every $x$ whence $\mbox{im}(1-p)\subseteq \ker p$.
Now take $x\in \ker p$. Then $x=x-px=(1-p)x$ belongs to $\mbox{im} (1-p)$. QED.
Remark: it is also useful to remember that the idempotents are exactly the diagonalizable operators with spectrum in $\{0,1\}$. If $p$ is nontrivial ($p\ne 0,1$), we have $\ker p$  the eigenspace of $0$, and $\mbox{im} p=\ker(1-p)$  the eigenspace of $1$. That is $\mbox{im} p=\{x\in V\,;\,px=x\}$, as opposed to $\ker p=\{x\in V\,;\,px=0$.

REPLY [2 votes]: First treating id$_U$=1 is correct. Because when you multiply a matrix by identity matrix, you got the matrix. So you can just ignore identity matrix during matrix multiplication, as you ignore 1 during multiplication.
Since $p^2 =p$ and $q$ is also a linear map defined by $q: U \rightarrow U$ then $q^2=q$.
But I believe the reasoning is the following computation similar to the previous one. The claim you made "because $q$ is also a linear map" justified the commutativity of addition for linear maps, a.k.a $p+q = q+p$:
$$q(q+p) = q(p+q) = q(\text{id}_U) \Rightarrow qp + q^2=q \Rightarrow q^2 =q.$$
Then following your own computation we're done.