We denote by $\lambda_{\text{max}}$ the largest of the eigenvalues of $J_n$ and $\lambda_{\text{min}}$ the smallest. Show that
$$\forall x \in \Lambda_n, \quad n\lambda_{\min} \leqslant \sum_{1 \leqslant i,j \leqslant n} J_n(i,j) x_i x_j \leqslant n\lambda_{\max}$$