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Defining $C_n(z) = \frac{z^m + z^{-m}}{2}$, the Chebyshev polynomials are defined by

$$T_n(C_1(z)) = C_n(z)$$ and are given by $T_1(z) = z, T_2(z) = 2z^2-1, T_3(z) = 4z^3-3z$, etc. Since for $z=e^{i\theta}$ we have $C_1(z) = \cos\theta$, they also satisfy

$$T_n(\cos\theta) = \cos(n\theta)$$

thereby generalizing the double angle trig identity.

The notes I'm reading also claim $$T_n(\text{tr } A/2) = \text{tr } (A^n/2)$$ for $A \in SL_2(\mathbb{C})$. Why does this follow?

Attempt: if $A= \begin{pmatrix} a&b \\ c &d \end{pmatrix}$ then choosing $z=\frac12(\sqrt{(a+d)^2-4}- (a+d))$ implies $C_1(z) = \text{tr }A/2$, but then $T_n(C_1(z)) = \frac{z^n+z^{-n}}{2}$ does not simplify as far as I see to $\text{tr} A^n/2$.

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    $\begingroup$ Note that the trace of $A$ is the sum of its eigenvalues and the product of its eigenvalues is by definition $1$. $\endgroup$ – WimC Dec 25 '18 at 18:09
  • $\begingroup$ @WimC thanks, that resolves it $\endgroup$ – Dwagg Dec 25 '18 at 18:57
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I found several sources with a proof, e.g., the paper by Francis Bonahon, Lemma $8$ on page $9$, using Cayley-Hamilton. Another interesting reference is the paper by Traina on trace polynomials for $SL_2(\Bbb{C})$.

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