Proving that if $A$ is a $8\times 8$ matrix over $\mathbb{R}$ and $A^3=A$, then $A$ is diagonalizable. 
If $A$ is a $8\times 8$ matrix over $\mathbb{R}$ and $A^3=A$ then prove that $A$ is diagonalizable.

I have got that the minimal polynomial of $A$ may be $x^3-x$ or $x$ or $x(x+1)$ or $x(x-1)$ in the 2nd case it is not possible in the 3rd and 4th cases the matrix will be the identity matrix or identity matrix multiplied by a scalar!
But I cannot reach any further.
 A: For any $v$, since $A(A-I)(A+I)v=0$, you get that:


*

*$v_{1}=(A^2+A)v$ is either zero or an eigenvector for eigenvalue $1$.

*$v_{-1}=(A^2-A)v$ is either zero or an eigenvector for eigenvalue $-1$.

*$v_0=(A^2-I)v$ is either zero or an eigenvector for eigenvalue $0$.


Now, note that:
$$v=\frac{1}{2}v_1+\frac{1}{2}v_{-1}-v_0$$
Why does this show that $A$ is diagonalizable?

In general, if $p(A)=0$ for some $p(x)=(x-a_1)(x-a_2)\cdots (x-a_k)$ for distinct $a_i$ (thus, not repeat roots) then $A$ can be diagonalized.
Specifically, if we define $p_i(x)=\prod_{j\neq i} (x-a_j)$ then we get that:
$$1=\sum_{i=1}^{n} \frac{1}{p_i(a_i)}p_i(x)$$
So if $v_i=p_i(A)v$ then $v_i$ is an eigenvector for eigenvalue $a_i$ and:
$$v = \sum \frac{1}{p_i(a_i)} v_i$$
A: Since the matrix $A$ is a zero of the polynomial $$p(x) = x^3 - x = x (x-1)(x+1)$$, the minimal polynomial of $A$ has to divide the polynomial $p$, but this means that $m$ is some combinations of the factors $x$, $(x-1)$, and $(x+1)$, but this implies that $A$ is diagonalisable.
QED.
