# 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!

$\begin{pmatrix} - N & - N + 1 & -N+2 & -N+3 &\ldots & 1(-2) & 1(-1) \\ - N + 1 & 2( - N + 1) & 2(-N+2) & 2(-N+3) &\ddots & 2(-2) & 2(-1) \\ - N + 2 & 2( - N + 2) & 3(-N+2) & 3(-N+3) &\ddots & 3(-2) & 3(-1) \\ - N + 3 & 2( - N + 3) & 3(-N+3) & 4(-N+3) &\ddots & 4(-2) & 4(-1) \\ \vdots & \vdots & \ddots & \vdots & \vdots \\ - 2 & 2(-2) & 3(-2) & 4(-2) &\ddots & ( - 1 + N)( - 2) & ( - 1 + N)( - 1) \\ - 1 & 2(-1) & 3(-1) & 4(-1) &\ldots & ( - 1 + N)( - 1) & N( - 1) \\ \end{pmatrix}$

• Do you just need to prove that the spectral radius is smaller than 1? This may be a much easier problem than finding an analytical expression for the eigenvalues. For instance, since the matrix has all negative entries, you can hope to use Perron-Frobenius theory to bound the maximum eigenvalue. – Federico Poloni Aug 22 '18 at 19:44

It's straightforward to show that this is the inverse of $1/(N+1)$ times the tridiagonal matrix $T_N$ with $-2$ on its main diagonal and $1$ on its super- and sub-diagonals.
Let $t_N$ be the characteristic polynomial of $T_N$. We have $t_0(x)=1$, $t_1(x)=x+2$, and by cofactor expansion $t_N(x)=(x+2)t_{N-1}(x)-t_{N-2}(x)$. That is, $t_N$ is related to the Chebyshev polynomials of the second kind by $t_N(x)=U_N(x/2+1)$. The roots of the Chebyshev polynomials are $\cos(\frac{k\pi}{N+1})$ for $k=1,\dots,N$, so the eigenvalues of the inverse of your $G$ are $\frac{2}{N+1}(\cos(\frac{k\pi}{N+1})-1)$. These have absolute value less than $1$ for $N\ge 3$.