Maximum eigenvalue of a covariance matrix of Brownian motion

Question

$$ A := \begin{pmatrix} 1 & \frac{1}{2} & \frac{1}{3} & \cdots & \frac{1}{n}\\ \frac{1}{2} & \frac{1}{2} & \frac{1}{3} & \cdots & \frac{1}{n}\\ \frac{1}{3} & \frac{1}{3} & \frac{1}{3} & \cdots & \frac{1}{n}\\ \vdots & \vdots & \vdots & \ddots & \frac{1}{n}\\ \frac{1}{n} & \frac{1}{n} & \frac{1}{n} & \frac{1}{n} & \frac{1}{n} \end{pmatrix}$$

How to prove that all the eigenvalues of $A$ are less than $3 + 2 \sqrt{2}$?

This question is similar to this one.

I have tried the Cholesky decomposition $A = L^{T} L$, where

$$L^{T} = \left(\begin{array}{ccccc} 1 & 0 & 0 & \cdots & 0\\ \frac{1}{2} & \frac{1}{2} & 0 & \cdots & 0\\ \frac{1}{3} & \frac{1}{3} & \frac{1}{3} & \cdots & 0\\ \vdots & \vdots & \vdots & \ddots & \vdots\\ \frac{1}{n} & \frac{1}{n} & \frac{1}{n} & \frac{1}{n} & \frac{1}{n} \end{array}\right)$$

then

$$(L^{T})^{-1}=\left(\begin{array}{ccccc} 1 & & & \cdots\\ -1 & 2 & & \cdots\\ & -2 & 3 & \cdots\\ \vdots & \vdots & \vdots & \ddots\\ & & & -(n-1) & n \end{array}\right)$$

$$A^{-1}=L^{-1}(L^{T})^{-1}$$

How to prove the eigenvalues of $A^{-1}$

$$\lambda_{i}\geq\frac{1}{3+2\sqrt{2}}$$

Further, I find that $A$ is the covariance matrix of Brownian motion at time $1, 1/2, 1/3, \ldots, 1/n$

Answer 1

In this answer I show that the largest eigenvalue is bounded by $5< 3 + 2\sqrt{2}$. I will first use the interpretation of this matrix as the covariance matrix of the Brownian motion at times $(\frac{1}{n},\dots, 1)$ (I reversed the order so that the sequence of times is increasing, which is more natural for me).

We have $A_{ij} = \mathbb{E} (B_{t_{i}} B_{t_j})$. The largest eigenvalue will be the supremum over the unit ball of the expression $\langle x, A x\rangle$, which is equal to $\sum_{i,j} A_{ij} x_{i} x_{j}$. This is equal to $\mathbb{E} (\sum_{i=1}^{n} x_{i} B_{t_{i}})^2$. In order to exploit the independence of increments of the Brownian motion, we rewrite the sum $\sum_{i=1}^{n} x_i B_{t_{i}}$ as $\sum_{i=1}^{n} y_{i} (B_{t_{i}} - B_{t_{i-1}})$, where $y_{i}:= \sum_{k=i}^{n} x_{k}$ and $t_0:=0$. Thus we have

$ \mathbb{E} (\sum_{i=1}^{n} x_{i} B_{t_{i}})^2 = \sum_{i=1}^{n} y_{i}^2 (t_{i}-t_{i-1}). $

The case $i=1$ is somewhat special and its contribution is $\frac{y_1^2}{n} \leqslant \sum_{k=1}^{n} x_{k}^2 = 1$. For the other ones we have $t_{i} - t_{i-1} = \frac{1}{(n-i+1)(n-i+2)}\leqslant \frac{1}{(n-i+1)^2}$. At this point, to get a nicer expression, I will reverse the order again by defining $z_{i}:= y_{n-i+1}$. So we want to estimate the expression

$ \sum_{i=1}^{n} \left(\frac{z_i}{i}\right)^2. $

We can now use use Hardy's inequality to bound it by $4 \sum_{i=1}^{n} x_{i}^2 =4$. So in total we get 5 as an upper bound, if I haven't made any mistakes.

Answer 2

Inspired by @Mateusz Wasilewski I find another method.

\begin{eqnarray*} \langle x,Ax\rangle & = & \langle Lx,Lx\rangle\\ & = & \sum_{i=1}^{n}u_{i}^{2} \end{eqnarray*}

where $u_{i}=\sum_{j=i}^{n}\frac{1}{j}x_{j}$.

\begin{eqnarray*} \sum_{i=1}^{n}u_{i}^{2} & = & \sum_{i=1}^{n}(\sum_{k=i}^{n}b_{k})^{2}\quad(\text{where} \ b_{k}=\frac{1}{k}x_{k})\\ & = & \sum_{i=1}^{n}(\sum_{k=i}^{n}b_{k}^{2}+2\sum_{k>j\geq i}b_{k}b_{j})\\ & = & \sum_{k=1}^{n}\sum_{i=1}^{k}b_{k}^{2}+2\sum_{j=1}^{n-1}\sum_{k=j+1}^{n}\sum_{i=1}^{j}b_{k}b_{j}\\ & = & \sum_{k=1}^{n}\frac{x_{k}^{2}}{k}+2\sum_{j=1}^{n-1}\sum_{k=j+1}^{n}b_{k}x_{j}\\ & = & \sum_{k=1}^{n}\frac{x_{k}^{2}}{k}+2\sum_{k=2}^{n}\sum_{j=1}^{k-1}b_{k}x_{j}\\ & = & \sum_{k=1}^{n}\frac{x_{k}^{2}}{k}+2\sum_{k=2}^{n}b_{k}z_{k-1}\\ & = & x_1^2 +\sum_{k=2}^{n}\frac{(z_{k}-z_{k-1})^{2}}{k}+2\sum_{k=2}^{n}\frac{(z_{k}-z_{k-1})}{k}z_{k-1}\\ & = & x_{1}^{2}+\sum_{k=2}^{n}\frac{z_{k}^{2}-z_{k-1}^{2}}{k}\\ & = & \sum_{k=1}^{n}\frac{z_{k}^{2}}{k}-\sum_{k=1}^{n-1}\frac{z_{k}^{2}}{k+1}\\ & = & \sum_{k=1}^{n-1}z_{k}^{2}(\frac{1}{k}-\frac{1}{k+1})+\frac{z_{n}^{2}}{n} \end{eqnarray*}