I am handling a problem of finding a unifom lower bound ( away from zero ) of the smallest eigenvalue of the following real symmetric matrix
$$A =\begin{pmatrix} n & n-1 & n-2 & \cdots & 1\\ n-1 & n-1 & n-2 & \cdots & 1 \\ n-2 & n-2 & n-2 & \cdots & 1 \\ \vdots & \ddots & \ddots & \ddots & \vdots \\ 1 & 1 & \cdots & \cdots & 1 \end{pmatrix}$$
I did some simulation (simulate n from 2 to 1500 ), it shows that the smallest eigenvalue of this matrix may have a lower bound around 0.25. So I try to prove it.
First, I found there may have some results related to the problem. Such as https://math.stackexchange.com/q/1330651
So I try to use
$$ \lambda_{min} \gt \sqrt{\frac{||A||_F^2-n||A||_E^2}{n(1-||A||_E^2/|det(A)|^{2/n})}} $$
or the result ( Theorem 1 ) in here. But only get a trivial lower bound zero.
Then I try to use min-max theorem to solve the following equivalent problem
$$\inf_{x\in \mathbb{R}^n,~\|x\|_2 = 1} x^{\top} A x$$
By some calculate, I think this problem is the following optimization
$$ \min_{x_i \in \mathbb{R}} x_1^2 + (x_1+x_2)^2 + \cdots (\sum_{i=1}^n x_i)^2~~s.t. ~~\sum_{i=1}^n x_i^2 = 1 $$
or at least find a lower bound of the object function. I tried the method of Lagrange multipliers, and the trigonometric functions to reformular the problem. But I got stuck because of the complicated calculation.
So, I wonder is there anyone may have some ideas about the problem?
Thanks!
Your observation is correct. The inverse of $A$ is a symmetric tridiagonal matrix: $$ A^{-1}=\pmatrix{ 1&-1\\ -1&2&-1\\ &-1&\ddots&\ddots\\ &&\ddots&\ddots&-1\\ &&&-1&2} $$ whose eigenvalues are known to be $\mu_j=4\sin^2\left(\frac{j-\frac12}{n+\frac12}\frac\pi2\right)$ for $j=1,\ldots,n$. So, the minimum eigenvalue of $A$ is $\dfrac1{\mu_n}=\left[4\sin^2\left(\frac{n-\frac12}{n+\frac12}\frac\pi2\right)\right]^{-1}$, which approaches $\frac14$ from the right when $n\to\infty$.