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 $(\text{det}A)^2=\det A$)?
 A: If $a + d = 1$, we have 
$a^2 + bc  = a$
$\iff a(1-d) + bc  = a$
$\iff -ad + bc = 0$
contradicting that det $A \ne 0$
From this we immediately get 
$ b = c = 0$
Furthermore 
$a, d \ne 0$ since $A$ is invertible and now $a^2 = a$ and $d^2 = d$ gives $a = d = 1$
A: Here is a linear algebra approach:
Note that $A^{2} = A \implies f(A) = 0$, where $f(x) =x(x-1)$.
This means that the minimal polynomial of $A$, denoted by $m_{A}(x)$, can only have three possibilities for it must be a divisor of $f(x)$ and those three possibilities are:
$m_{A}(x) = x, x-1$, and $f(x)$. If $m_{A}(x) = x$ or $x(x-1)$, then $\lambda = 0$ would be an eigenvalue of $A$ and that would mean $\det A =0$, which is a contradiction since $A \in GL_{2}(\mathbb{R})$. Hence, the only possibility is that $m_{A}(x) = x -1$.
We then have by the definition of a minimal polynomial that $0 = m_{A}(A) = A -I \implies A = I$, as it was promised. 
A: We have
\begin{aligned}
0=A^2-A
&=\pmatrix{a^2+bc-a&b(a+d-1)\\ c(a+d-1)&d^2+bc-d}\\
&=\pmatrix{a^2-(ad-bc)+ad-a&b(a+d-1)\\ c(a+d-1)&d^2-(ad-bc)+ad-d}\\
&=(a+d-1)A-(ad-bc)I\\
&=\left(\operatorname{tr}(A)-1\right)A-\det(A)I.
\end{aligned}
(Basically, we have reproved Cayley-Hamilton theorem for $2\times2$ matrices.) Therefore $\left(\operatorname{tr}(A)-1\right)A=\det(A)I$. Since the RHS is nonzero, $\operatorname{tr}(A)-1 $ must be nonzero too. In turn, $A=kI$ for some $k\ne0$. So, from $A^2=A$, we get $k^2I=kI$. Hence $k=1$ and $A=I$.
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 $a+d \ne 1$.
