[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.


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