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I'm working on the following problem.

What are the eigenvalues of a matrix with the property $M^3 = M$?

I can see the identity will satisfy this. I can also see that $Mx$ will be an eigenvector of $M^2$ but I don't know if that useful. I'm having trouble completely analyzing the problem. Thoughts?

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    $\begingroup$ If $Av=\lambda v$ then $\lambda^3 v = \lambda v$. If $v \neq 0$, what are the possible values of $\lambda$? $\endgroup$
    – copper.hat
    Sep 20, 2018 at 13:39

1 Answer 1

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Hint:

$M$ satisfies the polynomial $p(x)=x^3-x=0$

$p(x)$ can be factored into linear terms $p(x)= x(x-1)(x+1)$

Let $\pi(x)$ be the minimal polynomial of $M$ and by definition it is the monic polynomial of least degree such that $\pi(A)=0$ and any other polynomial $Q(x)$ with $Q(A) = 0$ is a (polynomial) multiple of $\pi(x)$.

So what are the possible options for $\pi(x)$?

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  • $\begingroup$ I haven't learned minimal polynomials. How do can I check $p(x)$ is the monic polynomial of least degree such that $p(A)$ is zero? $\endgroup$
    – yoshi
    Sep 20, 2018 at 16:42
  • $\begingroup$ Copper.hat's observation made sense to me, I'm just asking so I can learn something new. $\endgroup$
    – yoshi
    Sep 20, 2018 at 16:43
  • $\begingroup$ @yoshi $p(x)$ isn’t necessarily the minimal polynomial $\pi(x)$, but it constrains what the latter could be: $\pi(x)$ must be a divisor of $p(x)$. In this case, that gives you seven possibilities. $\endgroup$
    – amd
    Sep 20, 2018 at 21:24

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