 Solution

For an eigenvector $v$ with eigenvalue $\theta \in \{\lambda ,\mu \}$,By part (a), every vector orthogonal to $\mathbf{e}$ is a linear combination of such eigenvectors, so $B$ vanishes on ${\mathbf{e}}^{\perp }$. On the remaining eigenspace,The orthogonal projection onto $\mathrm{span}\{\mathbf{e}\}$ is $J/n$. It follows thatPut $\kappa =(d-\lambda )(d-\mu )/n$. For $i\ne j$, the identity $B=\kappa J$ readsand thereforeBecause $(A^2{)}_{ij}$ is the number of common neighbours, these two constants prove that $G$ is strongly regular. This is the converse half of the three-eigenvalue characterization of a connected strongly regular graph.