TITLE: Exercise 2.4 in Fulton and Harris
QUESTION [6 upvotes]: 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!

REPLY [3 votes]: 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.