It is well known that the characteristic polynomial of a bipartite graph is of the form
$\sum_{k=0}^n (-1)^kc_{2k} x^{2k}$ where $c_{2k} \geq 0$.
I can prove why there cannot be any odd powered coefficients in the characteristic polynomial but cannot find a way to prove that the coefficients of the remaining terms alternate sign.
I have tried working with Sachs Theorem which states that for a graph $G$ the coefficients $a_k$ are given by
$a_k(G) = \sum_{S \in L_k} (-1)^{\omega(S)}2^{c(S)}$
where $L_k$ denotes the set of subgraphs of G composed only of cycles and disjoint edges, $\omega(S)$ is the number of connected components in S and $c(S)$ is the number of cycles.
However, counting up all of the possible graphs and their contributions (particularly with different labellings) has proved tedious and tricky and I am wondering if there is a better way to approach the proof which I have not seen yet.
Thanks!