Upper Bound of Eigenvalues of Symmetric Real Matrix $A$

Question

Let $A_n=\begin{pmatrix} 1&\frac{1}{2}&\frac{1}{3}&\cdots&\frac{1}{n}\\ \frac{1}{2}&\frac{1}{2}&\frac{1}{3}&\cdots&\frac{1}{n}\\ \frac{1}{3}&\frac{1}{3}&\frac{1}{3}&\cdots&\frac{1}{n}\\ \vdots&\vdots&\vdots&\ddots&\vdots\\ \frac{1}{n}&\frac{1}{n}&\frac{1}{n}&\cdots&\frac{1}{n} \end{pmatrix}$. For each $n\in\mathbb{N}$, let $\lambda_0$ be one eigenvalue of $A_n$ ($\lambda_0$ is arbitrarily chosen)

Prove that $0<\lambda_0<3+2\sqrt{2}$.

A problem in a previous final exam. The number $3+2\sqrt{2}$ seems strange enough...

If $\alpha=(x_1, x_2,\cdots,x_n)^\mathrm{T}$ is an eigenvector of $A_n$ belonging to $\lambda_0$, we get $n$ equations: $$x_1+\frac{x_2}{2}+\cdots+\frac{x_n}{n}=\lambda_0x_1$$ $$\frac{x_1}{2}+\frac{x_2}{2}+\cdots+\frac{x_n}{n}=\lambda_0x_2$$ $$\vdots$$ $$\frac{x_1}{n}+\frac{x_2}{n}+\cdots+\frac{x_n}{n}=\lambda_0x_n$$ I tried to use inequalities to get a proof, but it doesn't work...There should be better approaches. Any advice or help appreciated.

Answer 1

For sake of removing this question from the "unanswered" tab, note that a proof was given in the question Upper bound for the eigenvalues of a matrix . The key observation is that the Cholesky decomposition of $A_n$ is $LL^T$, where $L$ happens to be the lower triangular part of $A_n$. It follows that if $D$ denotes the diagonal part of $A_n$, then $LL^T=A_n=L+L^T-D$. Hence $$ \rho(A_n)=\|A_n\|\le\|L\|+\|L\|+\|D\|=2\sqrt{\rho(A_n)}+1, $$ from which we can deduce that $\rho(A_n)\le3+2\sqrt{2}$.