Past exam of the mathematics course of the University of Cambridge / 2017 / ii / Paper 4 / 16H / Solution

For the Petersen graph, each vertex has three neighbours, adjacent vertices have no common neighbour, and distinct nonadjacent vertices have exactly one common neighbour. The entries of $A^2$ count length-two walks, so diagonal entries are three, adjacent entries zero, and other off-diagonal entries one. HenceThe connected graph has the simple eigenvalue three with eigenvector $1$. On its orthogonal complement $J=0$, so the remaining eigenvalues satisfy ${\lambda }^2+\lambda -2=0$ and are one or minus two. If their multiplicities are $r,s$, then $r+s=9$ and $\mathrm{tr}A=3+r-2s=0$. Thus