Hello,

Let $A$ be a $(n \times n)$ matrix over the complex numbers be such that $A^2 = I$. Show that $A$ is diagonalizable.

So I know that $\det (A^2 - \lambda I) = \det (I - \lambda I) = (1-\lambda)^n$ and the minimal polynomial, call it $n(x)$, of $A^2$ must divide $(1-\lambda)^n$.

Because 1 is the only complex number which solves the characteristic polynomial I also have $A^2(v) = \lambda v = 1 v = v$. $A^2 = I$ so $A^2$ should have rank $n$. So $n(x) = (1- \lambda)^n$?

The standard way to show that $A$ is diagonalizable is to show that it's minimal polynomial is equal to its characteristic polynomial. I was trying to prove that the characteristic polynomial of $AB$ must be divisible by either the minimal or the characteristic polynomials of A and/or B but I came up with a counterexample immediately.

My only other idea involves $A^2(v) = \lambda v$ implies $A(v) = \sqrt{\lambda} v$ but I also think that's false.

If you think of $A$ as ranging over values in $M_n(\mathbb{C}$, then the set of solutions for $A^2 - I$ is obviously $\{-I, I\}$ but now the trouble is in showing that no other matrices would satisfy that equation.

Does anyone have any advice?