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So I've been given a question that basically looks like this: Let $T:V \to V$ be a linear transformation with $T^3 = T \circ T \circ T = T$, Show that $T$ can only have eigenvalues from the set $\{-1,0,1\}$.

So I understand the question and how to calculate eigenvalues 'normally' but in this case the problem I have is what to actually write. I get the idea of the answer (I think) which is basically that only the values of $-1$, $0$ and $1$ can be cubed and get the same value back ($-1$ cubed is $-1$, $0$ cubed is $0$, etc.). I'm pretty much just not sure how I would 'prove' this to show what I need to show. Thanks in advance.

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2 Answers 2

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Let $\lambda,x$ be an eigenvalue/vector pair of $T$. Then $T x = \lambda x$. Then note that $\lambda x = Tx = T^3 x = T^2 (Tx) = T^2 (\lambda x) = \lambda T(Tx) = \lambda^2 T x = \lambda^3 x$.

Therefore $\lambda^3 = \lambda$ or equivalently $\lambda(\lambda^2-1)=0$ with solutions $\lambda = -1,0,1$.

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  • $\begingroup$ This make perfect sense, thanks a lot. $\endgroup$
    – Maximus
    Jan 26, 2021 at 16:01
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Your idea is correct. The underlying idea is that if $\lambda$ is an eigenvalue of $T$ then $\lambda^3$ is an eigenvalue of $T^3$. Can you prove this?

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  • $\begingroup$ I think so, thank you for the clarification. When put like that what I need to do seems a bit clearer. $\endgroup$
    – Maximus
    Jan 26, 2021 at 15:57

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