Let $E$ be a linear operator on $V$ for which $E^2 = E$, i.e., $E$ is a projection operator.
Let $U$ be the image of $E$, and let $W$ be its kernel.
a) If $u\in U$, then $E(u) = u$, i.e., $E$ is the identity mapping on $U$.
b) If $E\neq I$, then $E$ is singular, that is $E(v) = 0$ for some $v\neq 0$.
How can I prove these problems?
Answer for the second part.
If $E \neq I$ then for some $v \in V$, $E(v) = u$, where $v \neq u$. Now by $(a)$ we have $E(u) = u$. Thus $E(v - u) = E(v) - E(u) = u - u = 0$, where $v - u\neq 0$
(a) If $u \in U$, we can write $u = E(v)$ for some $v \in V$. Apply $E$ one more, to see $E(u) = E^{2}(v) = E(v) = u$.
(b) Otherwise, the kernel is zero, so $E$ is injective which leads to $E = I$ from $E^{2} = E$.
a) If $u\in U$, then there is $v\in V$ such that $E\left(v\right)=u$. Consequently $E\left(u\right)=E\left(E\left(v\right)\right)=$... you can apply the definition of a projector to finish to answer.
b) To answer it suffices to show that the kernel is not trivial. If $E=I$, it is clear that the kernel is the zero of $V$. So, suppose $E\neq I$, we know that $V$ is the direct sum of $U$ and $W$ (i.e. $V=U\oplus V$). For all $v\in V$, $v=u+w$ where $u\in U$ and $w\in W$. The end of the answer is not very difficult to establish.
