TITLE: Verify a Matrix Proof Given $A = A^2$
QUESTION [9 upvotes]: Problem
Given an arbitrary matrix $ A = A^2 $, prove that $ I - 2A = (I - 2A)^{-1}.$
Attempt
$$ I - 2A = (I - 2A)^{-1} $$
$$ (I - 2A)(I - 2A) = (I - 2A)(I - 2A)^{-1} $$
$$ (I - 2A)(I - 2A) = I $$
$$ (I - 2A)^2 = I $$
This is where I'm not sure if the distributive property can be applied to the following equation. Assuming it can be,
$$ (I - 2A)^2 = I $$
$$ I^2 - 4A + 4A^2 = I $$
$$ I^2 - 4A + 4A = I $$
$$ I^2 = I $$
$$ I = I $$
$$ [LHS] = [RHS] $$
Notes
Is it possible to prove this property is satisfied without using the distributive property as I did above? And what would be an actual example of such a matrix A that satisfies this property?
Solution to Proof
After reviewing and understanding multiple solutions, including that of @JMoravitz, I've come to the following solution.
Consider $ (I - 2A)^{2} $. We must first show that $ (I - 2A)^{2} = I$.
$$ (I - 2A)^2 = (I - 2A)(I - 2A) $$
$$ (I - 2A)^2 = I^2 - 4A + 4A^2 $$
$$ (I - 2A)^2 = I^2 - 4A + 4A $$
$$ (I - 2A)^2 = I^2 $$
$$ (I - 2A)^2 = I $$
Therefore, $ I - 2A $ is the inverse of $ I - 2A $, by the definition of matrix inversion.
An example of such a matrix that satisfies this property is the $ I_{2 \times 2} $ matrix, which is an idempotent matrix.
\begin{bmatrix} 
1 & 0 \\
0 & 1 
\end{bmatrix}

REPLY [2 votes]: Such a linear transformation, $A^2=A$, is called a projection.
For example, an orthogonal projection:
$$\begin{pmatrix}1&0&0\\0&1&0\\0&0&0\end{pmatrix}$$.
You seem to have all the elements of a proof that $I-2A=(I-2A)^{-1}$...