# Exercise 2.4 in Fulton and Harris

I'm trying to begin reading Fulton and Harris' Representation Theory and I'm having trouble with the following:

Exercise. Show that if we know the character $\chi_{V}$ of a representation $V$, then we know the eigenvalues of each element $g$ of $G$, in the sense that we know the coefficients of the characteristic polynomial of $g : V \rightarrow V$. Carry this out explicitly for elements $g \in G$ of orders $2, 3$, and $4$, and for a representation of $G$ on a vector space of dimension $2$, $3$, $4$.

I kind of get the hint that we should look at $\wedge^{k}V$ (more precisely on the eigenvalues of $g$ on $\wedge^{k}V$, which should be the products I pressume, but I don't think I understand the things well enough, so a detailed reponse would be really helpful for me at this point to get things clear. Many thanks in advance!

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If $g$ has eigenvalues $\lambda_1, ... \lambda_n$, then $$\chi_V(g^k) = \sum_{i=1}^n \lambda_i^k.$$
This sequence uniquely determines the eigenvalues. There are several ways to see this; the shortest is probably to observe that $$\sum_{k \ge 0} t^k \chi_V(g^k) = \sum_{i=1}^n \frac{1}{1 - t \lambda_i}$$
although this argument doesn't directly prove that in fact it suffices to take $k \le n$. For this, use Newton's identities.
One can also argue using exterior powers but it's not clear a priori that the character of an exterior power of $V$ can be determined from the character of $V$ (if you don't know that characters uniquely determine a representation); this is true but the shortest and most explicit argument I know also uses Newton's identities. – Qiaochu Yuan Feb 16 '12 at 4:20