$A\in M_3(\mathbb R)$ with $A^8=I$, then what can we tell about the degree of its minimal polynomial?

Question

What I understand that $A$ satisfies $x^8-1=0$, and as we know characteristic polynomial of $A$ is 3rd degree so characteristic polynomial is either $(x^2+1)(x+1)=0$ or$(x^2+1)(x-1)=0$

But I can not say anything about its minimal polynomial. Please help.

what about if $A=I$, then the minimal polynomial of $A$ is $t-1$ — Weijun Yin, Apr 16 '20 at 09:32
$x^4+1=(x^2-\sqrt2x+1)(x^2+\sqrt2x+1)$, so we have other degree $3$ polynomials like $(x^2-\sqrt2x+1)(x-1)$ — Qurultay, Apr 16 '20 at 09:35
Does this answer your question? Are there uncountably many $A\in M_3 (\mathbb {R})$ such that $A^8=I $? — Dietrich Burde, Apr 16 '20 at 09:42
@DietrichBurde: I think the proposed duplicate only makes a passing remark about possible minimal polynomials for this case. That Question was a multiple choice kind of exercise, and I don't see how any of the Answers there give an exhaustive response on the minimal polynomial. I will look again. — hardmath, Apr 16 '20 at 19:40
@hardmath We can easily read off the minimal polynomials of the (uncountably many) solutions to $A^8=I$ and together with Cayley-Hamilton this is already quite a lot. But yes, it is perhaps not an exact duplicate. It should be helpful as a hint, though. — Dietrich Burde, Apr 17 '20 at 08:49

score 0 · Answer 1 · answered Apr 16 '20 at 09:37

Hint: The minimal polynomial divides the characteristic polynomial by the Cayley-Hamilton theorem.

Reference:

Minimal polynomials and characteristic polynomials

Answer for $A\in M_3 (\mathbb {R})$ satisfying $A^8=I$:

Are there uncountably many $A\in M_3 (\mathbb {R})$ such that $A^8=I $?

hardmath · Answer 2 · 2020-04-17T16:32:16.143

As asked about matrix $A\in M_3(\mathbb R)$ satisfying $A^8=I$, "what can we tell about the degree of its minimal polynomial?" , the answer is that the degree of the minimal polynomial can be $1,2,$ or $3$. We show a bit more, roughly what minimal polynomials of $A$ are possible.

Since $A\in M_3(\mathbb R)$ and the minimal polynomial divides the characteristic polynomial (of degree $3$), the degree of the minimal polynomial is at most $3$. The minimal polynomial must divide $x^8 - 1$, since $A^8 = I$, and therefore the minimal polynomial also has no repeated roots.

It remains only to show that degrees $1,2,$ and $3$ can each be achieved. The case of degree $1$ requires a single eigenvalue (root of characteristic polynomial) of geometric multiplicity $3$, so take $A=I$ or $A=-I$.

The case of degree $2$ requires two distinct eigenvalues, one of geometric multiplicity $2$ and one of multiplicity $1$. For example:

$$ A = \begin{pmatrix} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & -1 \end{pmatrix} $$

Finally we consider the case of minimal polynomial having degree $3$. Recall that the companion matrix of a polynomial has characteristic polynomial equal to its minimal polynomial. Therefore any degree $3$ real polynomial dividing $x^8 - 1$ can be realized as the minimal polynomial of matrix $A$. From its real irreducible factors, as @Qurultay commented:

$$ x^8 - 1 = (x-1)(x+1)(x^2+1)(x^2-\sqrt 2 x+1)(x^2+\sqrt 2 x+1) $$

Thus we can choose either of its degree $1$ irreducible factors and any one of its three degree $2$ irreducible factors. Multiplying those two choices gives us a degree $3$ divisor of $x^8 - 1$, and that can be realized as the minimal polynomial (and characteristic polynomial) of $A$. For example, $(x+1)(x^2 + 1) = x^3 + x^2 + x + 1$ and:

$$ A = \begin{pmatrix} 0 & 0 & -1 \\ 1 & 0 & -1 \\ 0 & 1 & -1 \end{pmatrix} $$

$A\in M_3(\mathbb R)$ with $A^8=I$, then what can we tell about the degree of its minimal polynomial?

2 Answers2