Consider a matrix $A_n$ of size $n\times n$ $$ A_n = \begin{pmatrix} 2 & -1 & 0 & 0 & \dots & 0 & -1 \\ -1 & 2 & -1 & 0 & \dots & 0 & 0 \\ 0 & -1 & 2 & -1 & \dots & 0 & 0 \\ \dots & \dots & \dots& \dots& \dots& \dots & \dots\\ \dots & \dots & \dots& \dots& \dots& \dots & \dots\\ 0 & 0 & \dots & \dots & -1 & 2 & -1 \\ -1 & 0 & \dots & \dots & 0 & -1 & 2 \end{pmatrix} $$
As we know circulants have eigenvalues equal to the multiplication of the Fourier matrix by circulant's first column: $$ \begin{pmatrix} \lambda_1 \\ \lambda_2 \\ \lambda_3 \\ \vdots \\ \lambda_n \end{pmatrix} = \begin{pmatrix} 1 & 1 & 1 & \dots & 1 \\ 1 & w^{1\cdot 1}_n & w^{1\cdot 2}_n & \dots & w^{1\cdot (n-1)}_n\\ 1 & w^{2\cdot 1}_n & w^{2\cdot 2}_n & \dots & w^{2\cdot (n-1)}_n\\ \dots & \dots & \dots &\dots &\dots \\ 1 & w^{(n-1)\cdot 1}_n & w^{(n-1)\cdot 2}_n & \dots & w^{(n-1)\cdot (n-1)}_n\\ \end{pmatrix} \begin{pmatrix} 2 \\ -1 \\ 0 \\ \vdots \\0 \\ -1 \end{pmatrix} = \begin{pmatrix} 2 -1 -1 \\ 2 - w_n^{1} - w_n^{n-1} \\ 2 - w_n^{2} - w_n^{2(n-1)} \\ \vdots \\ 2 - w_n^{(n-1)} - w_n^{(n-1)(n-1)} \end{pmatrix} $$ Thus, $$ \lambda_k = 2 - w_n^k - w_n^{k(n-1)} = 2 - w_n^k - w_n^{-k} = 2 - e^{-\frac{2\pi i}{n}k} - e^{\frac{2\pi i}{n}k} = 2 - 2 \cos \frac{2\pi k}{n} = 4 \sin^2 \frac{\pi k}{n}, $$ So, $$\lambda_k = 4 \sin^2 \frac{\pi k}{n} \geq 0, \quad k=0,1,\dots, n-1$$ Therefore, our matrix is non-negative definite. Note that $\lambda_0 = 0$, hence matrix is singular.
Note, that $\|A_n\|_2 = \max_i |\lambda_i|$ only for Hermitian matrices!