Skip to main content
2 of 4
added 4 characters in body
user6818
  • 1.9k
  • 12
  • 20

Is there a way to simplify this apparently huge characteristic polynomial calculation?

Say I am given the $0/1$ adjacency matrix of an undirected graph. Also I am given a representation $\rho$ of some group $G$ and an orientation has been arbitrarily chosen along each edge. Let $E^{oriented}$ be the set of edges of the graph where for each edge one has chosen an orientation i.e choosing to denote a particular edge as $(a,b)$ instead of $(b,a)$ (its an independent choice at each edge) And I am given a map $s:E^{orient} \rightarrow \rho(G)$.

Now construct a matrix $B$ thought of as a $\vert V \vert \times \vert V \vert$ sized array of $dim(\rho)$ dimensional matrices such that if $(i,j) \in E^{orient}$ then the matrix at the $(i,j)$ array position is $= s((i,j))$ and if $(i,j) \not \in E^{orient}$ then that is $=s((j,i))^{-1}$.

Clearly $B$ is a $dim(\rho) \vert V \vert$ dimensional matrix.

  • Is there a short-cut or any simplification that can be done in calculating the characteristic polynomial of $B$ ? (than the obvious brute-force calculation of the determinant of a $dim(\rho) \vert V \vert$ dimensional matrix)
user6818
  • 1.9k
  • 12
  • 20