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Let $M$ be a $n\times n$ matrix such that $M^3=I$.Suppose that $Mv \neq v$ for any non zero vector $v$.Then which of the following is\are true?

1).$M+M^{-1}$ has real eigenvalues

The solution i tried is

Here it is given that $M^3=I$ so from here it is confirm that its minimal polynomial can be from $(x-1),(x^2+x+1) \;or\; (x-1)(x^2+x+1)$, but according to given condition $(x-1)\; and \;(x-1)\;(x^2+x+1)$ can't be minimal polynomial.So the only possibility is $(x^2+x+1)$

From above it is confirmed that roots are complex roots ,but still i have no idea how to prove $M+M^{-1}$ has real eigenvalue

please help!

Thankyou

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If $x^{2}+x+1$ is the minimal polynomial of $M$ then $M^{2}+M+I=0$. Also $M^{3}=I$ implies $M^{-1}=M^{2}$. Hence $$M^{-1}+M=M^{2}+M=-I$$.

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Let $\lambda$ be an eigenvalue of $M$. Then, since $M^3=\operatorname{Id}$, $\lambda^3=1$. But\begin{align}\lambda^3=1&\iff(\lambda-1)(\lambda^2+\lambda+1)=0\\&\iff\lambda^2+\lambda+1=0,\end{align}since $\lambda\neq1$. There are only two numbers $\lambda$ such that $\lambda^2+\lambda+1=0$: $-\frac12\pm\frac{\sqrt3}2i$. In each case, $\frac1\lambda=\overline\lambda$. If $v$ is an eigenvector corresponding to the eigenvalue $\lambda$, then$$(M+M^{-1}).v=\lambda v+\frac1\lambda v=\left(\lambda+\overline\lambda\right)v=2\operatorname{Re}(\lambda)v=-v.$$

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