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Let $A \in M_2(\mathbb R)$ a $2\times 2$ matrix with real coefficient, such that $A \ne I$ and $ A^3=I $ Then $\text{tr}(A)=-1$. What if we consider $M_n(\mathbb R)$? Is the statement still true?

I didn't manage to solve it, but I have a question: can we say $A$ is non singular? Indeed, $ A^3=I \Rightarrow AA^2=A^2A=I\Leftrightarrow A^{-1} = A^2. $

Is it right? How can we prove the statement?

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    Consider the 3x3 partitoned \begin{matrix}A & 0 \\ 0 & 1\end{matrix} , or the 4x4 \begin{matrix}A & 0 \\ 0 & A\end{matrix} 2012-09-15

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The minimal polynomial of $A$ must divide $X^3-1=(X-1)(X^2+X+1)$; since $A\neq I$ we have that the minimal polynomial of $A$ is $X^2+X+1$ (the degree of the minimal polynomial is smaller or equal to $2$).Because it has degree $2$, it must be equal to its characteristic polynomial, therefore $\operatorname{tr}(A)=-\text{ coefficient of }X=-1$

I don't think that it's true in general.For instance if $n=3$, the minimal polynomial of $A$ must be $X^3-1$ (why?), so $\operatorname{tr}(A)=0$.

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    I was referring to a theorem which say that: All eigenvalues of $A$ are the roots of the minimal polynomial of $A$. If $X^2+X+1$ is the minimal polynomial then $A$ has no real eigenvalue, but the characteristic polynomial of $A$ has degree $3$, and any polynomial of odd degree must have real root.2012-09-15
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You're right in that it is nonsingular. Otherwise, it would have a zero eigenvalue, so it would not be annihilated by a polynomial with a constant term.

$A$'s minimal polynomial divides $x^3-1$, so its eigenvalues are all third roots of unity.

If they were both $1$, $A$ would be identity (it can't have nontrivial Jordan blocks, because $x^3-1$ is squarefree), so at least one of them is nonreal.

$A$'s entries are real, so nonreal complex roots occur in conjugate pairs, and the sum of the two conjugate nonreal third roots of unity is $-1$.

In general, for an $n\times n$ matrix, the trace of such a matrix will be an integer of the form $n-3k$ with $0< k\leq \lfloor n/2\rfloor$ (and any such integer can be found as a trace of such a matrix, consider block matrices with diagonal blocks equal to $A$ or a $1\times 1$ matrix $(1)$ and $0$ elsewhere).