<s>[INST] If $M_A(x)=x^3-x$ So $M_{A^2}(x)=x^2-x$ If $M_A(x)=x^3-x$ So $M_{A^2}(x)=x^2-x$ Where $M$ is the minimal polynomial
$x^3-x=x(x^2-1)=x(x-i)(x+i)$ so $A$ is diagonalizable 
$A=P^{-1}\begin{pmatrix} 
0 & 0 & 0 \\
0 & i & 0 \\
0 & 0 & -i \\
\end{pmatrix}P$ 
$A^2=P^{-1}\begin{pmatrix} 
0 & 0 & 0 \\
0 & i & 0 \\
0 & 0 & -i \\
\end{pmatrix}^2P=P^{-1}\begin{pmatrix} 
0 & 0 & 0 \\
0 & -1 & 0 \\
0 & 0 & -1 \\
\end{pmatrix}^2P$ 
So $P_{A^2}(x)=x(x-1)^2$ but $A^2(A^2-1)=0$ so $M_{A^2}(x)=x(x-1)=x^2-x$
Is there another way to conclude it? 
 [/INST] From $M_A(x) = x(x-1)(x+1)$ we conclude that $A$ is diagonalizable and $\sigma(A) = \{0, -1, 1\}$. 
Hence $A^2$ is also diagonalizable and $\sigma(A)^2 = \sigma(A)^2 = \{0,1\}$. Therefore, the minimal polynomial $M_{A^2}$ has only linear factors and has zeroes $0, 1$.
The only option is $M_{A^2}(x) = x(x-1)$.
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