Conditions for generalized projection matrix of size (2x2)? My results seem incorrect... I am trying to derive the general conditions that must be imposed upon the real-valued entries of a $2 \times 2$ orthogonal projection matrix. However, I am coming to a conclusion that seems wrong and hope someone can point me in the right direction.
The prompt reads as follows:

Find conditions on $a,b,c,d \in \mathbb{R}$ that guarantee the matrix $\begin{pmatrix} a & b \\ c & d\end{pmatrix}$ defines a rank-1 orthogonal projection.

I begin by imposing the conditions that all projection matrices $P$ must fulfill, namely
(i) $P = P^T$
(ii) $P = P^2$
Applying condition (i),
$\begin{pmatrix} a & b \\ c & d\end{pmatrix} \overset{!}{=} \begin{pmatrix} a & c \\ b & d\end{pmatrix}$
I get $b=c$ and so I continue working with $\begin{pmatrix} a & c \\ c & d\end{pmatrix}$ and move on to condition (ii):
$\begin{pmatrix} a & c \\ c & d\end{pmatrix} \overset{!}{=} \begin{pmatrix} a & c \\ c & d\end{pmatrix}\begin{pmatrix} a & c \\ c & d\end{pmatrix}=\begin{pmatrix} a^2 +c^2 & ac + cd \\ ac +cd & d^2 + c^2\end{pmatrix}$
From that, I get three equations:
(1) $a = a^2 +c^2$
(2) $c=ac+cd=c(a+d)$
(3) $d=d^2 + c^2$
Further working out (2) by striking $c$ from each side, the system is then
(1) $a = a^2 +c^2$
(2) $1=a+d$
(3) $d=d^2+c^2$
I re-arrange (1) and (3) to get
(1) $c^2 = a - a^2$
(3) $c^2 = d - d^2$
and thus $a-a^2 = d-d^2$ from which I conclude that $a = d$. Returning to equation (2), I then get
(2) $1 = a+d = 2a$
and so $a = \frac{1}{2}$ and $d = \frac{1}{2}$.
Plugging either one of these into equation (1) or (3) then allows me to solve for $c$,
$\frac{1}{2} = \left ( \frac{1}{2} \right )^2+c^2$
or
$c^2 = \frac{1}{4}$
And thus $c=\pm \frac{1}{2}$.
So, according to these results, to be guaranteed an orthogonal projection matrix of rank-1, my matrix $P$ must be one of two possibilities (as indicated by $\pm$):
$P = \frac{1}{2} \begin{pmatrix} 1 & \pm 1 \\ \pm 1 & 1\end{pmatrix}$
This result is unsettling. First of all, I was expecting a broader range of possibilities for $a,b,c,d$ or at least for one or two variables. It seems odd that $a$ and $d$ are wholly constrained to one value and that $c$ only has two options.
Is this correct? It seems like there should be more here that is possible. Aren't matrices such as $\begin{pmatrix} 1 & 0 \\ 0 & 0 \end{pmatrix}$ or $\begin{pmatrix} 0 & 0 \\ 0 & 1 \end{pmatrix}$ also to be included here?
For example, when I apply $\begin{pmatrix} 1 & 0 \\ 0 & 0 \end{pmatrix}$ to a general vector $\begin{pmatrix} x \\ y \end{pmatrix}$, it's easy for me to see the orthogonal projection:
$\begin{pmatrix} 1 & 0 \\ 0 & 0 \end{pmatrix} \begin{pmatrix} x \\ y\end{pmatrix} = \begin{pmatrix} x \\ 0\end{pmatrix}$
But when I apply the resulting projection matrix from above to the same generalized vector, I get a result that does reveal orthogonal projection to me (using $c = + \frac{1}{2}$):
$\begin{pmatrix} \frac{1}{2} & \frac{1}{2} \\ \frac{1}{2} & \frac{1}{2}\end{pmatrix} \begin{pmatrix} x \\ y\end{pmatrix} = \frac{1}{2} \begin{pmatrix} x+y \\ x+y\end{pmatrix}$
How can that be an orthogonal projection? Isn't that simply a change in length along the same direction? Am I not understanding the fundamental concept?
Thanks for any help you might be able to provide.

According to this source, the projection matrix is given by
$ P = A (A^T A)^{-1} A^T $
So you can select $A$ to be any vector, and in particular a unit vector in $\mathbb{R}^2$, i.e.
$ A = [\cos \theta, \sin \theta ]^T $
The projection matrix parametrized by $ \theta $ is
$ P = \begin{bmatrix} \cos \theta \\ \sin \theta \end{bmatrix} \begin{bmatrix} \cos \theta && \sin \theta \end{bmatrix} = \begin{bmatrix} \cos^2 \theta && \cos \theta \sin \theta \\ \cos \theta \sin \theta && \sin^2 \theta \end{bmatrix}$
