Describe the set of all possible real projector matrices in $\mathbb{R}^{2 \times 2}$ So I started answering my own question as I went but there was a lot of tedious work (for me), and I couldn't find a list of projector matrices, so I figured I would leave this up for posterity. But please let me know if I missed any matrices or if there's an easier way of building the set (and of course if there's any errors). I posted my results as an answer below. Thank you.
 A: Hint:
consider that a projector has the eigenvalues $1$ and $0$ ( you can see: Eigenvalues of projection matrix proof). So any projector is represented by a matrix similar to a diagonal matrix with diagonal elements $1$ and $0$.
A: By definition, if $P \in \mathbb{R}^{2 \times 2}$ is a projector matrix then
$$P^2 = P$$    
and writing $P$ as   
$$P = \left(\begin{array}{cc}a & b\\c & d\end{array}\right)$$  
we could solve the system from $P(P - I) = 0$  
$$\left(\begin{array}{cc}a & b\\c & d\end{array}\right) \left(\begin{array}{cc}a-1 & b\\c & d-1\end{array}\right) = 0$$  
In which case we get the system of equations    
$$ \begin{array}  
aa(a-1) + bc = 0 \\ 
ab + b(d-1) = 0 \\ 
c(a-1) + dc = 0 \\ 
bc + d(d-1) = 0
\end{array} \quad \to \quad
\begin{array}   
aa^2- a + bc = 0 \\ 
ab + bd - b = 0 \\ 
ac - c + dc = 0 \\ 
bc + d^2 - d = 0
\end{array}   
$$  
Subtracting the last equation from the first we get   
$$a^2 - d^2 = a - d \implies (a-d)(a+d) = a-d \implies a =d \quad \text{or} \quad a = 1-d$$

Case 1: $a = 1-d$  
Plugging into equation 2 we get 
$$  
(1-d)b + b(d-1) = 0
 $$    
Which implies equation 2 is satisfied for all $b$. Plugging into equation 3 we get a similar result for $c$. But multiplying equations 2 and 3 we get a common factor of $bc$, yet the product $eqn. 1 \times eqn. 2$ needs to equal zero. This implies $bc = 0$ (note the remaining factor has no solution which means $bc = 0$ is the only way to satisfy the system). So now, if $b$ and $c$ are arbitrary subject to their product equaling zero we get (eqn 1)
$$   
a(a-1) = 0  
\implies \boxed{a = 0,1} \implies \boxed{d = 1,0} \quad \text{respectively}
$$    
our solution set in this case is
$$  
P_1 = \left(\begin{array}{cc}1 & b\\c & 0\end{array}\right)  \quad \text{and}  
\quad 
P_2 = \left(\begin{array}{cc}0 & b\\c & 1\end{array}\right)\,,  \quad 
\text{for all} \, \, b\cdot c = 0
$$     
that is, either $b$ or $c$ has to be zero.   
For example, $P_2$ with $b = 0$, multiplied by a vector in $\mathbb{R}^2$ scales the $x$ component by factor $c$, adds the result $xc$ to the $y$ component of the vector, and maps the result to the $y$-axis. Here's what it looks like in desmos. Note that $P_2$ becomes an orthogonal projector when $c = 0$.  
Here's what the $P_1$ set looks like in desmos (note that the projector is orthogonal when $c = 0$).

Case 2: $ a = d$  
Applied to equation 3   
$$c(d-1) + dc = 0 \implies c(2d-1) = 0 \implies c = 0 \quad \text{or} \quad 
d = \frac{1}{2}$$  
Inspecting equation 1 we see that for $c = 0$ we have   
$$a(a-1) = 0 \tag{$\star$}$$    
A quick analysis of $(\star)$: $a = d = c = 0 \implies b = 0$ which is the trivial case of the zero matrix. If $a = d = 1$ then equation 2 implies $b = 0 = c$ which is the other trivial case of the identity matrix. Done with $(\star)$. 
But nontrivially, for $d = 1/2$ we find via elementary algebra that our only restriction is that $bc = 1/4$. That is   
$$  
P_3 = \left(\begin{array}{cc}1/2 & b\\c & 1/2\end{array}\right) \quad bc = 1/4 
$$  
It projects onto hyperbolas it looks like, but I'm not sure how to make it work.

Summary: The only 2D real projector matrices are the identity matrix, the zero matrix, and the sets  
$$  
P_1 = \left(\begin{array}{cc}1 & b\\c & 0\end{array}\right)\, ,  
\quad 
P_2 = \left(\begin{array}{cc}0 & b\\c & 1\end{array}\right)\,,  \quad 
\text{for all} \, \, bc = 0\, , \\ \text{and}  \quad  
P_3 = \left(\begin{array}{cc}1/2 & b\\c & 1/2\end{array}\right) \quad bc = 1/4 
$$
A: Let's look for all rank $1$ projectors, because rank $0$ and rank $2$ are obvious.
If $P$ is such a projector, then its characteristic polynomial is $X^2-X$. In particular, $\det P=0$ and $\operatorname{trace}P=1$. Thus the matrix must have the form
$$
P=\begin{bmatrix} a & b \\ c & 1-a \end{bmatrix}
$$
If we square it, we get
$$
P^2=\begin{bmatrix}
a^2+bc & b \\
c & bc+(1-a)^2
\end{bmatrix}
$$
so the relations become
$$
\begin{cases}
a(1-a)-bc = 0 \\[4px]
a^2+bc=a \\[4px]
bc+(1-a)^2=1-a
\end{cases}
$$
which however are equivalent to one another. Thus the condition is just
$$
bc=a-a^2
$$
Actually, the check about $P$ being idempotent is not necessary. The condition that the trace is $1$ is satisfied and $\det P=0$ implies $bc=a-a^2$. Now the characteristic polynomial is $X^2-X$ and Hamilton-Cayley provides $P^2=P$.
