Q:

Prove that identity matrix is the only idempotent $n x n$ matrix that is invertible.

I do get that let's say 
$A=(A*A)$ 
$A^{-1} * A = A^{-1} *(A*A)$
$(A^{-1}*A)=(A^{-1}*A)A$
Then $I=I*A$, therefore $I=A$
Are those correct? Is there another or proper way to prove this?

A:

Your proof is great, and completely correct.  Personally, I object to the use of $*$ as denoting matrix multiplication, but that's more an opinion that anything.

Another version of the proof that I like is as follows:
Let's say that $A$ is invertible with $A = A^2$. Then we have
$$
A^2 - A = 0 \implies\\
A(A - I) = 0 \implies\\
A^{-1}[A(A - I)] = A^{-1}[0] \implies\\
A-I = 0 
$$
So, we must have $A - I = 0$, which is to say that $A = I$.
Why do I like this better?  I suppose the approach generalizes nicely to arbitrary polynomials and the consideration of minimal polynomials, but really I just think factoring is neat.

