Example for Matrix with $A^3=A$ that is not diagonalizable

Question

I am working on an exercise where I am supposed to prove of disprove that $A\in\mathbb K^{n\times n}$ with $A^3=A$ is diagonalizable.
My work: $$A^3=A\iff A^2(1-A)=0\Rightarrow \mu_A(X)=X\ \vee\ \mu_A(X)=X(X-1)\ \vee\ \mu_A(X)=X^2(X-1)\ \vee\ \mu_A(X)=X^2$$ where $\mu_A$ denotes the minimal polynomial of $A$
Because $A$ is diagonalizable iff its minimal polynomial factors into distinct linear factors, $A$ with $\mu_A(X)=X^2 $ or $\mu_A(X)=X^2(X-1)$ cannot be diagonalizable.
Can somebody please provide examples that satisfy $A^3=A$ and $\mu_A(X)=X^2 $ or $\mu_A(X)=X^2(X-1)$? I fail to come up with any. Thanks!

Answer 1

There's a silly mistake in your argument. $A^3=A$ means $\mu_A(X)$ divides $X^3-X$. The latter polynomial factors as $$X^3-X=X(X-1)(X+1),$$ and not as $X^2(X-1)$ as you claimed.

Therefore, if $\text{char} \mathbb{K} \neq 2$, $\mu_A(X)$ factors as a product of distinct linear factors and hence $A$ is diagonalizable.

Answer 2

Here is an example of a non-diagonalizable matrix $A\in M_2(\Bbb F_2)$ which satisfies $A^3=A$, namely $$ A=\begin{pmatrix} 1 & 1 \cr 0 & 1 \end{pmatrix}. $$ We have $A^3=A$, because $3=1$ in $\Bbb K=\Bbb F_2$.