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let $M$ be an $n \times n$ matrix with real entries such that $M^3=I$. Suppose $Mv \neq v$ for any non zero vector $v$. Then which of the following statements is/are true?
A. $M$ has real eigenvalues
B. $M+M^{-1}$ has real eigenvalues

We have given, $Mv \neq v$ for any non zero vector $v$. That means, $M$ cannot have $1$ as an eigenvalue. For all $n \geq 3$, $p(x)=x^3-1$ is a polynomial that satisfies the given matrix $M$, and then $1$ is always the eigen value.
I am not able to generalize the given statements to any conclusion.Is my approach going in the right direction or not? Also I would like to know whether there is any condition on $n$ for such a matrix.

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Hint:

  1. If $p(M)=0$ with $p(X)=X^3-1$, and if $1$ is not an eigenvalue, then the only possible eigenvalues are$\ldots$

  2. Now if $\lambda$ is an eigenvalue of $M$, then one easily sees that $\ldots$ is an eigenvalue of $M+M^{-1}$.

  3. Finally, when $\lambda$ is one of the possible eigenvalues found at Hint 1, the corresponding eigenvalue according to Hint 2 happens to be real because$\ldots$


Regarding the value of $n$, the only caveat is that when $n=1$, the condition "$M^3=$Id but $1$ is not an eigenvalue" simply cannot happen. Otherwise, the reasoning is always valid, and there are examples from $n=2$ onwards.

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  • $\begingroup$ Got it. @arnaud. Thanks $\endgroup$ – Atul Anurag Sharma May 11 '18 at 2:33
  • $\begingroup$ @Atul you're welcome. $\endgroup$ – Arnaud Mortier May 11 '18 at 2:39
  • $\begingroup$ What can we say about $n$?. Is there any condition on $n$. $\endgroup$ – Atul Anurag Sharma May 11 '18 at 2:40
  • $\begingroup$ @Atul the only condition is $ n> 1 $. If $ n=1 $ then $M^3=1$ implies that $ 1 $ is an eigenvalue, so there are no examples where the conditions are met. Starting from $ n=2 $ there are examples among rotation matrices. $\endgroup$ – Arnaud Mortier May 12 '18 at 0:30
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Like you noted, if $p = X^3 -1$ then $p(M) = 0$ and therefore the minimal polynomial of $M$ will divide $p$. Since the roots of $p$ are the 3rd rooths of unity, the minimal polynomial of $M$ will have (some of) these as roots. This means that the eigenvalues of $M$ will be a subset of $G_3$. Now, since

$$ M + M^{-1} = M + M^{-1}\cdot M^3 = M + M^2 = ev_M(X^2+X) $$

this matrix will have eigenvalues $\mu^2 + \mu$ with $\mu$ an eigenvalue of $M$. But since $\mu$ will be a third root of unity, $\mu^2 = \mu^{-1} = \bar{\mu}$. Hence the eigenvalues of $M + M^{-1}$ are of the form

$$ \mu^2 + \mu = \mu^{-1} + \mu = \bar{\mu} + \mu \in \mathbb{R} $$

Regarding the first point, since $1$ cannot be an eigenvalue, $M$ must have complex eigenvalues because $G_3 \setminus \{1\} \subseteq \mathbb{C} \setminus \mathbb{R}$.

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