# [0x78392345]An Interesting Problem About Bounds of A Matrix’s Eigenvalues

For a real symmetric matrix $A= \left[\begin{array}{ccccc} 1 & {1\over{2}}&{1\over{3}}&...&{1\over{n}}\\{1\over{2}}&{1\over{2}}&{1\over{3}}&...&{1\over{n}}\\{1\over{3}}&{1\over{3}}&{1\over{3}}&...&{1\over{n}}\\...&...&...&...&...\\{1\over{n}}&{1\over{n}}&{1\over{n}}&...&{1\over{n}}\end{array} \right]$, the eigenvalues of it will be positive and smaller or equal to $3+2\sqrt{2}$

We could enhance the bound to be 5 actually, but that is pretty much good.

Here I will present a proof linked with Analysis, but the techniques involved would be more or less falling in the categories of high school competitions.

We wish to prove $x^{T}Ax < 5x^Tx$ and let $\alpha_k=\sum^{j=k}_{j=1}{x_j}$.

Observe the RHS: $x^{T}Ax$ could be rewritten as

$RHS={\sum^{j=n-1}_{j=1}{({{1\over {j}}-{1\over{j+1}}}){\alpha_k}^2}}+{{1\over {n}}{{\alpha_n}^2}}$.

By Cauchy-Schwartz, we get an estimate: ${1\over {n}}{\alpha_k^2} \leq {\sum^{j=k}_{j=1}{x_j^2}}$.

And now we end up proving

$RHS={\sum^{j=n-1}_{j=1}{{{1\over {j(j+1)}}}{\alpha_k}^2}}+{{1\over {n}}{{\alpha_n}^2}} \leq 4x^{T}Ax$.

To estimate the remaining part, we will estimate it by some positive $\beta_k$, a undecided parameters for which ${\alpha_k^2} \leq {\sum^{k}_{j=1}{{x_j^2}\over{{\beta_j}}}}{\sum^{k}_{j=1}{\beta_j}}$ holds.

And after plugging in,

after switching sum signs we could get $RHS={\sum^{j=n-1}_{j=1}{{{1\over {j(j+1)}}}{\alpha_k}^2}} \leq {\sum_{j=1}^{n-1}(\sum_{k=j}^{n-1}{{{{\beta}_1+...+{\beta}_k}}\over{k(k+1)}}) {{{x_j}^2}\over{{{\beta}_j}^2}}}$.

We will now prove ${1\over{\beta_j}}\sum_{k=j}^{n-1}{{{{\beta}_1+...+{\beta}_k}}\over{k(k+1)}} < 4$(*)

pick $\beta_m = {m\over{\sqrt{m+1}}}-{{m-1}\over{\sqrt{m}}}$ and so ${{\beta}_1}+...+{\beta}_k={k \over{\sqrt{k+1}}}$

hence the LHS of (*) will be ${1\over{\beta_j}}\sum_{k=j}^{n-1}{1\over{(k+1)^{3/2}}}$

and notice that ${\sum_{k=j}^{\infty}{1\over{(k+1)^{3/2}}}} \leq {2\over{\sqrt{j}}}$, our new estimate would be ${1\over{\beta_j}}\sum_{k=j}^{n-1}{1\over{(k+1)^{3/2}}} <{2\over{{\sqrt{j}{\beta_j}}}}$ It is easy to show ${{\sqrt{j}}{\beta_j}} > {1 \over {2}}$ hence the conclusion.

I do not know any other proof out there yet, just this proof out there—seems to be a lot of computation.