Prove that the eigenvalues of a random matrix of this form, are invariant regardless of the value of the exponent $s$.

Question

Consider the sequence:

$$\frac{\sum _{k=1}^1 \frac{1^s}{k^s}}{1^s},\frac{\sum _{k=1}^2 \frac{2^s}{k^s}}{2^s},\frac{\sum _{k=1}^3 \frac{3^s}{k^s}}{3^s},...,\frac{\sum _{k=1}^n \frac{n^s}{k^s}}{n^s} \;\;\;\;\;\;\;\;\;(1)$$

This of course simplifies to the Harmonic numbers:

$$\sum _{k=1}^1 \frac{1}{k^s},\sum _{k=1}^2 \frac{1}{k^s},\sum _{k=1}^3 \frac{1}{k^s},...,\sum _{k=1}^n \frac{1}{k^s}$$

Some Mathematica code to try it out:

s = 3;
nn = 12;
a = Table[Sum[(n^s/k^s), {k, 1, n}]/n^s, {n, 1, nn}]
HarmonicNumber[Range[nn], s]

But focus on the summand in the numerator in sequence $(1)$ for a moment.

Let $A$ be a matrix with arbitrary (random) entries:

$$A=\left( \begin{array}{ccccccc} a_{1,1} & a_{1,2} & a_{1,3} & a_{1,4} & a_{1,5} & ... & a_{1,k} \\ a_{2,1} & a_{2,2} & a_{2,3} & a_{2,4} & a_{2,5} & ... & a_{2,k} \\ a_{3,1} & a_{3,2} & a_{3,3} & a_{3,4} & a_{3,5} & ... & a_{3,k} \\ a_{4,1} & a_{4,2} & a_{4,3} & a_{4,4} & a_{4,5} & ... & a_{4,k} \\ a_{5,1} & a_{5,2} & a_{5,3} & a_{5,4} & a_{5,5} & ... & a_{5,k} \\ \vdots & \vdots & \vdots & \vdots & \vdots & \ddots & \vdots \\ a_{n,1} & a_{n,2} & a_{n,3} & a_{n,4} & a_{n,5} & ... & a_{n,k} \end{array} \right)$$

Multiply elementwise the arbitrary matrix $A$ with the numerator of the summand in $(1)$, so that you get:

$$A_{\text{Harmonic}}=\left( \begin{array}{ccccccc} \frac{1^s}{1^s}a_{1,1} & \frac{1^s}{2^s}a_{1,2} & \frac{1^s}{3^s}a_{1,3} & \frac{1^s}{4^s}a_{1,4} & \frac{1^s}{5^s}a_{1,5} & ... & \frac{1^s}{k^s}a_{1,k} \\ \frac{2^s}{1^s}a_{2,1} & \frac{2^s}{2^s}a_{2,2} & \frac{2^s}{3^s}a_{2,3} & \frac{2^s}{4^s}a_{2,4} & \frac{2^s}{5^s}a_{2,5} & ... & \frac{2^s}{k^s}a_{2,k} \\ \frac{3^s}{1^s}a_{3,1} & \frac{3^s}{2^s}a_{3,2} & \frac{3^s}{3^s}a_{3,3} & \frac{3^s}{4^s}a_{3,4} & \frac{3^s}{5^s}a_{3,5} & ... & \frac{3^s}{k^s}a_{3,k} \\ \frac{4^s}{1^s}a_{4,1} & \frac{4^s}{2^s}a_{4,2} & \frac{4^s}{3^s}a_{4,3} & \frac{4^s}{4^s}a_{4,4} & \frac{4^s}{5^s}a_{4,5} & ... & \frac{4^s}{k^s}a_{4,k} \\ \frac{5^s}{1^s}a_{5,1} & \frac{5^s}{2^s}a_{5,2} & \frac{5^s}{3^s}a_{5,3} & \frac{5^s}{4^s}a_{5,4} & \frac{5^s}{5^s}a_{5,5} & ... & \frac{5^s}{k^s}a_{5,k} \\ \vdots & \vdots & \vdots & \vdots & \vdots & \ddots & \vdots \\ \frac{n^s}{1^s}a_{n,1} & \frac{n^s}{2^s}a_{n,2} & \frac{n^s}{3^s}a_{n,3} & \frac{n^s}{4^s}a_{n,4} & \frac{n^s}{5^s}a_{n,5} & ... & \frac{n^s}{k^s}a_{n,k} \end{array} \right)$$

Prove or disprove that the eigenvalues of $A_{\text{Harmonic}}$ are the same regardless of the value of the exponent $s$.

In the program below I try to demonstrate that this is the case. Only problem is that Mathematica does not know how to sort complex numbers in a consistent order and therefore it sometimes gives non zero differences. So the program only demonstrates the weaker statement that the sum of the differences of the eigenvalues is zero.

The sum of the differences of the eigenvalues of two different matrices:

$$\sum\limits_{n=1}^{n=m} \lambda(1)_n-\lambda(2)_n = 0$$

is zero. But as I said, I am asking to prove that the eigenvalues actually are equal all the time.

Mathematica:

Clear[t, nn, n, k];
nn = 12
A = Table[
   Table[RandomReal[{-10, 10}] + RandomReal[{-10, 10}]*I, {k, 1, 
     nn}], {n, 1, nn}];
aa = Chop[Table[s = RandomReal[{-10, 10}] + RandomReal[{-10, 10}]*I;
   a = Table[Table[A[[n, k]]*(n^s/k^s), {k, 1, nn}], {n, 1, nn}];
   N[Eigenvalues[a]], {nn, 1, 7}]]
bb = Chop[Table[s = RandomReal[{-10, 10}] + RandomReal[{-10, 10}]*I;
   b = Table[Table[A[[n, k]]*(n^s/k^s), {k, 1, nn}], {n, 1, nn}];
   N[Eigenvalues[b]], {nn, 1, 7}]]
Print["Eigenvalues of the matrices are equal:"]
Chop[aa - bb]
Print["The sum of the differences of the eigenvalues is zero:"]
Chop[Table[Total[(aa - bb)[[m]]], {m, 1, Length[aa - bb]}]]