Prove that matrix $A$ diagonalizable if $A^2=I$ using characteristic polynomial 
Prove that the matrix $A$ is diagonalizable if $A^2=I$ using characteristic
  polynomial

I saw an answer that used the minimal polynomial of $A$. Can that be proven without using minimal polynomial? In my university we didn't learn minimal polynomial and therefore we cannot use it in our tasks and exams. 
We didn't get this particular question for homework but I just wanted to know if I can use this claim in my tasks and exams.
 A: Let us do it over the real numbers - it has been noted in a comment that this does not hold in characteristic 2.
Let $v$ be any vector of the underlying vector space $V$. Then
$$\tag{eq}
v = \dfrac12 \left((A + I) v - (A - I) v\right).
$$
(BTW, this does not hold when you cannot divide by $2$, i.e., in characteristic $2$.)
Moreover each vector $(A - I)v$ is an eigenvector of $A$ relative to the eigenvalue $-1$, as
$$
A ((A -I) v) = (A^2 - A) v = (I - A)v = -(A -I) v,
$$
and each vector $(A + I)v$ is an eigenvector of $A$ relative to the eigenvalue $1$, as
$$
A ((A + I) v) = (A^2 + A) v = (I + A)v = (A + I) v.
$$
Because of (eq), each vector is a sum of an element of the image $U$ of $A - I$ and an element of the image $W$ of $A + I$. Choose a basis of $U$ (which we have seen is made of eigenvectors of $A$ with respect to the eigenvalue $-1$) and a basis of $A$ (which we have seen is made of eigenvectors of $A$ with respect to the eigenvalue $1$), and you will obtain a basis of $V$ made of eigenvectors of $A$.
We have used the fact that $U \cap W = \{ 0 \}$. In fact if $z \in U \cap W$, then $z = A z = - z$, and thus $2 z = 0$, so that $z = 0$, again because we can divide by $2$.
