In the book "Graph Theory" - by Bondy and Murty exercise 1.1.22 b) ii) wants me to prove that $$det(J-(1+\lambda)I_n) = (1+\lambda-n)(1+\lambda)^{n-1}$$ where $J$ is the $n$ x $n$ matrix consisting only of $1$'s.
I think this is wrong by a factor of $(-1)^n$. This part of the exercise exists essentially only as a step to calculate the eigenvalues of a complete graph. For the eigenvalues the sign of the characteristic polynomial does not matter. But to me that doesn't change the fact that what they want you to show is incorrect. Can someone confirm this is in fact an error, and I'm not just making a mistake in calculating the determinant?
For $n=1$, \begin{align*} \det(J - (1+\lambda)I_1) = \det(-\lambda) &= -\lambda, \text{ but} \\ (1+\lambda-1)(1+\lambda)^{1-1} &= \lambda. \end{align*} So yes, the given formula is wrong.