If $A$ is idempotent then $A$ is similar to a diagonal matrix with only $0$'s and $1$'s on the diagonal.

Question

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.

Answer 1

If $A^2=A$, then the equality should hold for each of the Jordan blocks. In more detail, $A=SJS^{-1}$. Then $$ SJS^{-1}=A=A^2=SJ^2S^{-1}, $$ and then $J^2=J$. Now you can compare the individual Jordan blocks. In a Jordan block $J_1$, that is not diagonal and with eigenvalue $\alpha$, the $1,2$ entry is $1$. In $J_1^2$, the $1,2$ entry is $2\alpha$. So you would need $2\alpha=1$, which cannot happen here because $\alpha$ is either $0$ or $1$.

Answer 2

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.