If $A$ is idempotent then $A$ is similar to a diagonal matrix with only $0$'s and $1$'s on the diagonal. I am trying to use Jordan normal form to show that if $A^2 = A$ then it is similar to a diagonal matrix with only $0$'s and $1$'s. I've proved that the eigenvalues of $A$ have to be either $0$ or $1$ so we know the diagonal elements of the JNF have to be $0$ or $1$. How do we know that none of the off-diagonals above are $1$?
I don't full understand JNF - I've only learned about it in terms of elementary divisors and the minimal and characteristic polynomials, but a lot of resources online talk about it in terms of eigenspaces which is confusing.

You don’t really need the JNF for this. Since $A^2=A$, its minimal polynomial divides $x^2-1$, therefore there are only three possibilities for the minimal polynomial: $x$, $x-1$ and $x(x-1)$. In each case the polynomial is a product of linear factors, hence $A$ is diagonalizable.
