Calculating the diagonal entry of the adjugate of the Laplacian matrix of a complete graph.

Question

I am trying to directly calculate the adjugate of the Laplacian of the complete graph from which I know is given by $\tau(G) = n^{n-2}$ where $\tau(G)$ is the number of spanning trees of $K_n$.

I calculated it indirectly but my textbook (Norman Biggs, Algebraic Graph Theory chapter 6) says that it is a simple determinant manipulation on $nI-J$ where $J$ is the matrix of ones, but I cannot see this manipulation. The Laplacian is $$ \begin{bmatrix} n-1 & -1 & \cdots & -1 \\ -1 & n-1 & \cdots & -1 \\ \vdots & \vdots & \ddots & \vdots \\ -1 & -1 & \cdots & n-1 \end{bmatrix}. $$ The adjugate of the diagonal entry is the determinant of the Laplacian with the 1st row and column removed (w.o.l.o.g.).

The way I did it was by considering the eigenvalues of the $K_n$ with $0$ once and $n$, $n-1$ times and then used the formula which we derived in class from the Matrix Tree theorem giving $\tau(G) = \frac{1}{n}\prod$ non zero eigenvalues of Laplacian of $G$.

Answer 1

Found a way to do this by considering the adjugate entry of the first row first column, $M= nI_{n-1} -J_{n-1} $. Then, it is clear that $j = (1, \ldots, 1)$ is an eigenvector with eigenvalue $n-1-(n-2) = 1$ and other $n-1$ eigenvectors are of the form $x_i = (1, 0, \ldots, 0, -1, 0, \ldots, 0)$ where $-1$ is in the $i$th row. There is $n-2$ of them and they all have an eigenvalue $n$. So w.r.t. the eigenbasis of $M$, $$ M=\operatorname{diag}(1,n,n, \ldots, n) $$ which has a determinant $n^{n-2}$ as required.