Question
a) Prove that a matrix $A\in M_n(\mathbb{C})$ satisfying $A^3=A$ can be diagonalized.
b) Does the statment in (a) remain true if one replaces $\mathbb{C}$ by an arbitrary algebraically closed field $F$?
Answer
First of all, we can see that the matrix $A$ satisfies the polynomial $p(X)=X(X-1)(X+1)$. Since the minimal polynomial has to divide $p(X)$, we can conclude that the minimal polynomial of $A$ has no repeated root. This would yield that $A$ can be diagonalized. However, I couldn't answer the part (b). Actually, I do not know what the crucial part $\mathbb{C}$ plays in part (a) either. Thanks a lot in advance...
The answer is no; the matrix $$\begin{bmatrix} 1 & 1 \\ 0 & 1\end{bmatrix}$$ with entries in the algebraic closure of $\mathbb F_2$ (for example) is not diagonalizable but satisfies $A^3=A$. Its characteristic and minimal polynomial is $(x-1)^2$ which is equivalent to $x^2-1$ in characteristic $2$.
“Algebraically closed” in this case is a bit of a red herring; the fact that $\mathbb C$ has characteristic zero, hence $x^3-x$ truly splits into linear factors, is the reason why it works over $\mathbb C$. In the case above, $$x^3-x = x(x^2-1) = x(x-1)^2$$ so your minimal polynomial might not split into distinct linear factors.