If $M_A(x)=x^3-x$ So $M_{A^2}(x)=x^2-x$

Question

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?

Answer 1

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 $M_{A^2}(x) = x(x-1)$.

Answer 2

First you can plug in $A^2$ to see whether it satisfies the polynomial. Indeed:

$$(A^2)^2 - A^2 = A^4 - A^2 = A A^3 - A^2 = AA - A^2=0$$

Now check that $A^2$ doesn't satisfy the linear factors $x$ and $x-1$.