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Let $B$ be a $2\times 2$ matrix with integer entries and determinant $1$. Show that if $λ\in \mathbb{C}$ is a root of the characteristic polynomial and $|λ|= 1$, then there is $n \ge 1$ such that $λ^n= 1$. Can you help me start?

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    $\begingroup$ Deleting a question after getting an answer is kinda rude to the person who composed the answer. Therefore we have a very strong norm not to do that. Like, never. May be as a new user you didn't know about it, but if you stop to think about it for a moment, you will surely see why that is so. The reward to the asker comes from exposure of their work to other viewers. And, we don't really own the posts we make here. Read the fine print of legalese :-) $\endgroup$ – Jyrki Lahtonen Oct 23 '17 at 6:33
  • $\begingroup$ Having said that I refrain from voting here. I'm fairly sure this question has appeared earlier, and hence should possibly be closed as a duplicate. I don't have the time to look for an earlier incarnation now, sorry. $\endgroup$ – Jyrki Lahtonen Oct 23 '17 at 6:34
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    $\begingroup$ @JyrkiLahtonen :-/ Check revision history please. $\endgroup$ – Simply Beautiful Art Oct 24 '17 at 14:43
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    $\begingroup$ Do not vandalise your posts. $\endgroup$ – Daniel Fischer Oct 24 '17 at 15:05
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By the complex conjugate root theorem, the two roots of the characteristic polynomial must be conjugate pairs. So the polynomial looks like $p(t) = (t-\lambda)(t-\bar{\lambda})$ for some $\lambda \in \mathbb{C}$ with $|\lambda|=1$.

Additionally, one can directly show that $p(t) = \det(tI - B) = t^2 - \text{tr}(B) + \det(B)$. Thus by matching coefficients, $$2 \Re (\lambda) = \lambda + \bar{\lambda} = \text{tr}(B).$$

Recall $|\lambda| = 1$. Since the trace is an integer and $\Re(\lambda) \in [-1,1]$, we must have $\Re(\lambda) \in \{-1, -\frac{1}{2}, 0, \frac{1}{2}, 1\}$. From here, you can explicitly list all possible $\lambda$, and show that each is a root of unity.

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  • $\begingroup$ I am confused about n ≥1, where is the index coming from? $\endgroup$ – user491981 Oct 23 '17 at 6:00
  • $\begingroup$ @gohan: Perhaps by index you mean exponent. In any case the powers $n$ that work here are small (because the characteristic polynomial is only degree $2$). $\endgroup$ – hardmath Oct 24 '17 at 3:35