In one step of solving a difficult problem, I would like to know the largest eigenvalue of a matrix with this pattern: $$A_n = \begin{bmatrix} 0 & 0 & 0 & 0 &\dots & 0 \\ 0 & 1 & 1 & 1&\dots & 1 \\ 0 & 1 &2 &2 &\dots &2\\ \vdots & \vdots & \vdots & \vdots & \ddots & \vdots \\ 0 & 1 & 2 & 3 & \dots& n-1 \end{bmatrix}$$

In other words, we can "peel" the "L" shaped layers from upper left corner and all layer consist of the same entry.

This matrix is symmetric, so all eigenvalues are real. It is of rank $n-1$, and numerical experiment demonstrates that it is positive semidefinite. I want to know the largest eigenvalue in terms of $n$, or have a lower bound on the largest eigenvalue in terms of $n$. I performed numerical experiment on various $n$ and it seems like we can quickly get $\lambda_{max}\approx 0.4n^2$ for large $n$. However, I don't know how to derive this result. I tried to look at the characteristic polynomial, but I could not find a pattern.

So anyone has an idea about deriving a (lower) bound on the largest eigenvalue of this matrix? Thanks!

matrix(n,n,i,j,min(i,j)-1)] has characteristic polynomial $\sum_{k=0}^{n-1} (-1)^k {n-1+k \choose n-1-k} x^{n-k}$, and nonzero roots $-2\sum_{k=1}^{n-1} k \cos(2\pi jk/(2n-1))$ for $0<j<n$, with $j=1$ producing the maximal root. $\endgroup$ – Noam D. Elkies Sep 4 '18 at 1:50