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Let $A\in M_3(\mathbb{R})$ be such that $A^8=I$, Then :

$(1)$ The minimal polynomial of $A$ can only be of degree $2$.

$(2)$ The minimal polynomial of $A$ can only be of degree $3$.

$(3)$ either $A=I$ or $-I$

$(4)$ There are uncountably many $A$ satisfying the above

Here's my approach: An annihilating polynomial of $A$ is $p(x)=x^8-1 = (x-1)(x+1)(x^2+1)(x^4+1)$ and we know that $m_A(x)|p(x)$ and also $m_A(x)|\chi_A(x)$ and $\chi_A(x)$ is of degree $3$ so $m_A(x)$ must be at most degree $3$.

Let $A= \begin{pmatrix} 0 & -1 & 0 \\ 1 & 0 & 0 \\ 0 & 0 & 1 \end{pmatrix}$ (as given by Ravi Fernando in the comments) then $m_A(x)=(x-1)(x^2+1)$ so Option $(1)$,$(2)$,$(3)$ are false

4th option is true because we can take all the matrix which are similar to $A$

Lucas
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    The minimal polynomial might be $(x-1)(x^2+1)$ for instance. – lhf Sep 11 '20 at 21:02
  • Yes! Yes! I was confused and it arises because I was trying to find the minimal polynomial for any $A$ satisfying $A^8 = I$ – Lucas Sep 11 '20 at 21:15
  • can you give me an example of a real matrix $A$ which satisfy $(x-1)(x^2+1)$ other than identity – Lucas Sep 11 '20 at 21:19
  • the last row came up $(0,0,1)$ when computed second power – Lucas Sep 11 '20 at 21:40
  • @JCAA I think the last entry should be $1$, not $-1$. – Ravi Fernando Sep 11 '20 at 21:41
  • @RaviFernando even if the last entry is $1$ it doesn't matter. In the second power of the matrix, the last row will be $(0,0,1)$ – Lucas Sep 11 '20 at 21:46
  • What do you mean by complex similar? does it mean if A ~ B then there exists $P\in Gl_n(\mathbb{C})$ such that $A = PBP^{-1}$. We want a matrix $A\in M_n(\mathbb{R})$ such that $A^2 = -I$ – Lucas Sep 11 '20 at 22:22
  • There doesn't exists any real $3\times 3$ matrix $A$ such that $A^2 = -I$ since $det(A)^2 = -1$ cannot happen. So the minimal polynomial must be of degree at most 2 – Lucas Sep 11 '20 at 22:59
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    Right, we cannot have $A^2 = -I$. But there do exist $3 \times 3$ matrices $A$ with minimal polynomial $(x-1)(x^2 + 1)$, such as $$A = \begin{pmatrix} 0 & -1 & 0 \ 1 & 0 & 0 \ 0 & 0 & 1 \end{pmatrix}.$$ Note here that both $A-I$ and $A^2 + I$ are nonzero, but their product is zero. – Ravi Fernando Sep 12 '20 at 06:04
  • Ohh wow! I didn't see that. How did you construct such example? – Lucas Sep 12 '20 at 08:04
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    Note that you didn't factor $p(x)$ into irreducibles over $\mathbb R$, since $x^4+1 = (x^2+\sqrt 2 x+1)(x^2-\sqrt 2 x + 1)$. – Christoph Sep 12 '20 at 08:21

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