How to find the analytical representation of eigenvalues of the matrix $G$? I have the following matrix arising when I tried to discretize the Green function， now to show the convergence of my algorithm I need to find the eigenvalues of the matrix $G$ and show it has absolute value less than 1 for certain choices of $N$. 
Note that the explicit formula for entry $(i,j)$ is $-i(N+1-j)$ when $i\le j$ and it is symmetric, so we can get the formulas for $i>j$ by interchanging $i$ and $j$ in the $i\le j$ case. 
Any one has any ideas about how to find the analytical representation of eigenvalues of the matrix $G$, i,e, the eigenvalues represented by $N$? Thank you so much for any help!
$\begin{pmatrix}
 - N & - N + 1 & -N+2 & -N+3 &\ldots & 1(-2) & 1(-1) \\
 - N + 1 & 2( - N + 1) & 2(-N+2) & 2(-N+3) &\ddots & 2(-2) & 2(-1) \\
 - N + 2 & 2( - N + 2) & 3(-N+2) & 3(-N+3) &\ddots & 3(-2) & 3(-1) \\
 - N + 3 & 2( - N + 3) & 3(-N+3) & 4(-N+3) &\ddots & 4(-2) & 4(-1) \\
 \vdots & \vdots & \ddots & \vdots & \vdots \\
 - 2 & 2(-2) & 3(-2) & 4(-2) &\ddots & ( - 1 + N)( - 2) & ( - 1 + N)( - 1) \\
 - 1 & 2(-1) & 3(-1) & 4(-1) &\ldots & ( - 1 + N)( - 1) & N( - 1) \\
\end{pmatrix}$
 A: Eigenvalues of matrix $G$ are
$$\lambda_k(G)=-\tfrac{1}{4}(N+1)\sin^{-2}\left(\frac{k\pi}{2(N+1)}\right)\,,\quad k=1,\ldots,N\,.$$
To prove the statement, we start by writing elements of matrix $G$ as $g_{ij}=ij-(N+1)\min(i,j)$.
Next, we define matrix $L$ as a lower triangular matrix with all elements in the lower triangle equal to one, i.e. $l_{ij}=1$ for $i\geq j$ and $l_{ij}=0$ for $i<j$, and we define vector $\mathbf{1}$ as a vector of ones.
This enables us to write
$$G=(L\mathbf{1})(L\mathbf{1})^T-(N+1)LL^T = L(\mathbf{1}\mathbf{1}^T-(N+1)I)L^T\,.$$
Since $L^{-1}$ is bidiagonal matrix with $1$ on the diagonal and $-1$ on the subdiagonal, we hoped inverting $G$ would result in a matrix with more zero elements.
Using Sherman-Morrison formula to invert $\mathbf{1}\mathbf{1}^T-(N+1)I$, we obtain
$$G^{-1}=\frac{-1}{N+1}L^{-T}(I+\mathbf{1}\mathbf{1}^T)L^{-1} = \frac{-1}{N+1}\begin{bmatrix}\hphantom{-}2 & -1 & & 0\\-1 & \ddots & \ddots\\ & \ddots & \ddots & -1\\0 & & -1 & \hphantom{-}2\end{bmatrix}\,.$$
This matrix is a tridiagonal Toeplitz matrix, and there is analytical expression for eigenvalues of such matrices, see this link for example. In our case 
$$\lambda_k(G^{-1}) = \frac{-2}{N+1}+\frac{2}{N+1}\cos\left(\frac{k\pi}{N+1}\right) = \frac{-4}{N+1}\sin^2\left(\frac{k\pi}{2(N+1)}\right)\,,\quad k=1,\ldots,N\,.$$
Now, our statement follows from eiganvalues of $G$ being reciprocals of eigenvalues of $G^{-1}$.
