conditions for idempotence in $2 \times 2$ matrix Let $A=\begin{bmatrix} a&b\\c&d\\\end{bmatrix}$.  I'm looking for the conditions such that $A^2=A$.  So I start calculating...
$$A^2=\begin{bmatrix} a&b\\c&d\\\end{bmatrix}\begin{bmatrix} a&b\\c&d\\\end{bmatrix}=\begin{bmatrix} a^2+bc&ab+bd\\ac+cd&bc+d^2\\\end{bmatrix}=\begin{bmatrix} a&b\\c&d\\\end{bmatrix}=A$$
So there are things we can find...we have
1)  $ab+bd=b\rightarrow b(a+d)=b \rightarrow (a+d)=1 \rightarrow d=(1-a).$  This tells me that $Tr(A^2)=Tr(A)=1$ under this condition of idempotence.  Calculation on $ac+cd=c$ produces the same result, that $d=(1-a)$
2)  $a^2+bc=a \rightarrow bc=a-a^2 \rightarrow bc=a(1-a)\rightarrow bc=ad\rightarrow ad-bc=0 \rightarrow det(A)=0$.  This leads me to that idempotent matrix is singular.  Calculation
So it would seem that these were shown to be true by direct calculation.  Yet the Identity matrix $I_2$ is clearly idempotent and yet $Tr(I_2)=2$ and $det(I_2)=1$  So why is this the case?  How does the direct calculation not support the identity case?  Are there any easier ways to show the above while including the case of $I_2$?
 A: As $A^2=A$ the eigenvalues of $A$ are $0$ or $1$. We have three cases:
Case 1: $\lambda_1=\lambda_2=0$. Then the characteristic Polynomial of $A$ is $X^2=0$. Then $A^2=0$ and hence $A=A^2=0$.
Case 2: $\lambda_1=\lambda_2=1$. Then $A$ is invertible, and we get $A^2=A \Rightarrow A=I$.
Case 3: $\lambda_1=0, \lambda_2=1$. Then $A$ is diagonalizable and hence
$$A=\begin{bmatrix} x&y\\z&t\\\end{bmatrix}\begin{bmatrix} 1&0\\0&0\\\end{bmatrix}\begin{bmatrix} x&y\\z&t\\\end{bmatrix}^{-1}=\frac{1}{xt-zt} \begin{bmatrix} x&y\\z&t\\\end{bmatrix}\begin{bmatrix} 1&0\\0&0\\\end{bmatrix}\begin{bmatrix} t&-y\\-z&x\\\end{bmatrix}=... $$
A: By Cayley-Hamilton we know that $A^2-Tr(A)A+\det(A)I_2=0$. Hence $A^2=A$ means $A=I_2$, $A=0$, or $\det(A)=0$ and $Tr(A)=1$.
A: If $A$ is an $N\times N$ matrix such that $A^{2}=A$, then $A$ is a projection matrix. Every $y\in\mathbb{R}^{N}$ can be written as
$$
      y= Ay +(I-A)y .
$$
Notice that $A(Ay)=(Ay)$ and $A(I-A)y=0$. That is, $A$ is $I$ on the range of $A$ and is $0$ on the range of $I-A$. So there is a basis $\{ y_{1},\ldots,y_{N}\}$ such that the transformed matrix $A$ is 0 everywhere except on the diagonal, where it will have 1's or 0's. The case for $\mathbb{R}^{2}$ is a special case.
