Prove an idempotent invertible 2x2 matrix in general linear group $\text{GL}_2(\mathbb{R})$ must be the identity The general linear group of degree $2$ over $\mathbb{R}$ (all $2 \times 2$ invertible real matrices with matrix multiplication), is a group. In any group $G$, if $a \in G$ is idempotent i.e. $aa=a$, then
$$a=ae=a(aa^{-1})=(aa)a^{-1}=(a)a^{-1}=e$$
In $\text{GL}_2(\mathbb{R})$, the identity is $e=I_2= \begin{bmatrix}1&0 \\ 0 &1\\ \end{bmatrix}$. I want to show without using the above result that if $A \in \text{GL}_2(\mathbb{R})$ is idempotent, then $A=I_2$.
Suppose $A=AA$, so
$$
\begin{bmatrix} a & b\\ c & d\\ \end{bmatrix}=A=AA= \begin{bmatrix} a & b\\ c & d\\ \end{bmatrix}\begin{bmatrix} a & b\\ c & d\\ \end{bmatrix} = \begin{bmatrix} a^2+bc & b(a+d)\\ c(a+d) & d^2+bc\\ \end{bmatrix}
$$
Thus, $a^2+bc=a,\, d^2+bc=d,\, b(a+d)=b,\, c(a+d)=c$. Using the last two equations, we have $c(a+d-1)=0=b(a+d-1)$. There are three cases: $b=0$, $c=0$, or $a+d=1$. 
I used the additional fact that $A$ is invertible and $ad \neq bc$ to show that $A=I_2$ for $b=0$ and $c=0$. Notice that if $\text{tr}A=a+d=1$, then it is impossible to have $A=I_2$, but I have had no success with finding a contradiction. What breaks if $\text{tr}A=1$?
Perhaps we can use that $\text{det}A=ad-bc=1$ for any invertible idempotent matrix (since $(\text{det}A)^2=\det A$)?

If $a+d=1$, then
$$A = \begin{bmatrix} a & b\\ c & 1-a\\ \end{bmatrix}$$
and $A = A^2$ implies
$$\begin{bmatrix} a^2+bc & b\\ c & 1-a\\ \end{bmatrix}
  = \begin{bmatrix} a & b\\ c & a^2+bc-2a+1\\ \end{bmatrix}$$
and we must have $$a^2-a+bc = 0$$
But this implies $\det A = 0$; which is a contraction, since $A$ is invertible. So $a+d \ne 1$.
