# Eigenvalues of M

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?

• 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 $M + M^{-1} = -I$. – астон вілла олоф мэллбэрг Jun 15 '18 at 6:08

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

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