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Let M be square matrix of order n with real entries Satisfying $M^3=I$ and $Mv\neq v$ for any non-zero vector $v$.Then Which of the followings are true?

$1.$ M has real eigen-values?

$2.$ $M+M^{-1}$ has real eigen-values?

$3.n$ is divisible by 2.

$4.n$ is divisible by 3.

$Mv\neq v$ for any non-zero vector $v$ means 1 is not an eigenvalue of M.Hence $x^2+x+1=0$ is minimal polynomial of M.Hence 2 is correct option.But I am not able to understand why 1st and 4th options are correct?

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    $\begingroup$ The first statement definitely cannot be true, since if $\lambda$ is an eigenvalue of $M$ then $Mv= \lambda v \implies v = \lambda^3 v$ so $\lambda^3 = 1$ but $\lambda \neq 1$ so $\lambda$ is necessarily complex. The second is right since $M + M^{-1} = -I$. $\endgroup$ – астон вілла олоф мэллбэрг Jun 15 '18 at 6:08
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  1. $M^3=I$ so $M^3-I=(M-I)(M^2+M+I)=0$ but for hypothesis you have that $M-I $ is invertibile and so: $M^2+M+I=0$ By contraddiction if there exists a real eighenvalue $\lambda$ of M then , if $v\neq0$ is his eighenvector, you have that $(M^2+M+I)v=(\lambda^2+\lambda+1)v=0$ and so $\lambda^2+\lambda+1=0$ But this polinomial equation not have solutions in $\mathbb{R}$.

  2. $M^2+M+I=0$ and you can multiply both members for $M^{-1}$ :

$M+I+M^{-1}=0$

So

$M+M^{-1}=-I $ that is diagonalizzable oviously.

  1. What you said is correct.

By contraddiction if $ n=deg(det(M-\lambda I) $ is not divisibile for 2 then the polinomial have necessary a real root because any polinomial of degree odd have al least a root in $\mathbb{R}$

  1. It is false because there exists a $2x2$- real matrix $M$ such that $M^3=I$ and $M-I$ is invertibile. An example can be

$\begin{bmatrix}-\frac{1}{2}&\frac{\sqrt{3}}{2}\\ -\frac{\sqrt{3}}{2}&-\frac{1}{2}\end{bmatrix}$

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