Given $A^2 = I$ show $A$ is diagonalizable I have seen many solutions online involving the minimal polynomial but I was wondering if there is another way to prove the above without using the minimal polynomial?
 A: Well...you can sort of do the work by hand. You know that
$$
A^2 = I \\
A^2 - I = 0 \\
(A-I)(A+ I) = 0
$$
so either $A-I$ or $A + I$ cannot be full-rank, because if they were, their product would be...but it's definitely not (unless the dimension is zero...)
Suppose that $A - I$is not full-rank. Then it has a kernel, ie., a vector $v$ with $$
(A-I)v = 0 \\
Av - Iv = 0 \\
Av - v  = 0 \\
Av = v
$$
so it has an eigenvector for eigenvalue 1. ... and you keep going in this direction to reduce the dimension, while retaining the same $A^2 = I$ property (might be a little tricky, but I don't think so), and when you're done, the set of eigenvectors you've produced constitutes a basis wrt which $A$ is diagonal. 
A: One solution is based on the Jordan normal form. Suppose that $A^k=I$ for a positive integer $k.$ Let $A'=P^{-1}AP$ be the Jordan normal form of $A$ and let $B$ be an $m\times m$ Jordan block of $A'.$ Then $B^k=I_m.$ This is only possible if $m=1,$ so we conclude that $A'$ is diagonal.
A: Note that 
$$I = \frac{I+A}{2} + \frac{I-A}{2}$$
(no brainer)
and 
$$A \cdot \frac{I+A}{2} = \frac{I+A}{2}\\
A \cdot \frac{I-A}{2} = -\frac{I-A}{2}$$
since $A^2 = I$.
For every vector $v$ we get 
$$v = \frac{I+A}{2}v + \frac{I-A}{2}v= v_1 + v_{-1}$$
with $A v_1 = v_1$, $Av_{-1} = - v_{-1}$
Therefore $V_{1} + V_{-1}$ is the whole space. It is clear that the intersection of them is $(0)$. Therefore $V = V_1 \oplus V_{-1}$.
Now take a basis of $V$ as a union of a basis of $V_1$ and a basis of $V_{-1}$. In this basis $A$ has a diagonal matrix.
