This is a simple problem of linear algebra. One without knowing graph theory may solve it. I am missing a small easy logic.
Description: Let $G$ be a graph with $n$ vertices and $G^c$ is its complement in the complete graph $K_n$. The Laplacian matrices of $G$ and $G^c$ are given by $L(G)$ and $L(G^c)$. The following relation is well known. $$L(G) + L(G^c) = L(K_n) = nI_n - J_n$$ $I_n$is identity matrix of order $n$ and $J_n$ is a matrix with all entrees equal to $1$.
Let us assume that the eigenvalues of $L(G)$ are $\lambda_1 \ge \lambda_2 \ge \dots \lambda_{n-1} \ge 0$. Here let us assume $G$ is a connected graph then $\lambda_{n-1} > 0$.
Question: Prove that the eigenvalues of $L(G^c)$ are $(n - \lambda_{n - 1}, n - \lambda_{n-2} \dots , n - \lambda_1 , 0)$
The eigenvalues of $I_n$ are all $1$. As each element of $J_n$ are $1$, its trace is $n$ and determinant is $0$. So eigenvalues will be $n$ with multiplicity $1$ and $0$ with multiplicity $n-1$. Here I am not getting a small and easy logic to prove the result. Please Explain.
Thank you for your help.
If $\lambda_{n-1}>0$ then the eigenspace associated to eigenvalue $\lambda_0=0$ of $L(G)$ consists solely of multiples of $e:=(1,1,\dots,1)^T$. The eigenvectors to the eigenvalues $\lambda_1\dots \lambda_{n-1}$ are thus orthogonal to $e$.
Now take the an eigenvector $v_i$ of $L(G)$ associated to $\lambda_i$, $i\in\{1\dots n-1\}$. Then we obtain $$ L(G^c)v_i = nv_i - J_nv_i -L(G)v_i = (n-\lambda_i)v_i $$ since $J_nv_i = 0$ due to $e^Tv_i=0$. Hence, $v_i$ is an eigenvector of $L(G^c)$ as well with corresponding eigenvalue $n-\lambda_i$.