Gre practice problem on Linear Transformations The GRE Practice book includes this problem (#37).

Let $V$ be a finite-dimensional real vector space and let $P$
  be a linear transformation of $V$ such that $P^2 = P$.
  Which of the following must be true? 
I. $P$ is invertible.
II. $P$ is diagonalizable.
III. $P$ is either the identity transformation or the zero transformation.
(A) None
(B) I only
(C) II only
(D) III only 
(E) II and II

Here are my (modest) thoughts on the problem so far.
Choice I) Not necessarily true because $P$ could be the zero transformation.
Choice II) I really don't know anything about this. What is the significance of diagonalizable transformations?
Choice III) $P$ certainly could be the identity or zero transformations, but I don't know how to prove or disprove the existence of other valid $P$s
 A: Such a transformation is called a projection. The only invertible projection is the identity: if $P^2=P$ and $P$ is invertible, then 
$$
P=P^{-1}P^2=P^{-1}P=I.
$$
A projection $P$ is always diagonalizable. Two ways I can see of seeing it: note that the eigenvalues of $P$ are $0$ and $1$ (unless $P=0$ or $P=I$, which are the easy cases). 


*

*The decomposition $V=PV\oplus (I-P)V$ suggests that if you choose a basis of $V$ made of bases of $PV$ and $(I-P)V$, you will have a basis of eigenvectors. 

*Alternatively, look at the Jordan form of $P$ and conclude that it has to be diagonal. 
Nontrivial projections: here are infinitely many, in $M_2(\mathbb R)$:
$$
 \begin{bmatrix}t&\sqrt{t-t^2}\\\sqrt{t-t^2}&1-t\end{bmatrix},\ \ t\in[0,1]. 
$$
Or:
$$
\begin{bmatrix}1&x\\0&0\end{bmatrix},\ \ x\in\mathbb R.
$$
A: your thoughts on choice $1$ is right.
it always helps to cook up a few small examples satisfying the conditions, like
$$
 \begin{bmatrix}1& 1\\0 &0\end{bmatrix} 
$$
and
$$
 \begin{bmatrix}1& 0\\0 &0\end{bmatrix} 
$$
