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 $I_2$?
Best Answer
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
$$A=\begin{bmatrix} x&y\\z&t\\\end{bmatrix}\begin{bmatrix} 1&0\\0&0\\\end{bmatrix}\begin{bmatrix} x&y\\z&t\\\end{bmatrix}^{-1}=\frac{1}{xt-zt} \begin{bmatrix} x&y\\z&t\\\end{bmatrix}\begin{bmatrix} 1&0\\0&0\\\end{bmatrix}\begin{bmatrix} t&-y\\-z&x\\\end{bmatrix}=... $$