Showing that a Hermitian matrix can have eigenvalues that correspond to arbitrary numbers does not prove the Hilbert-Polya conjecture, does it?

Question

I read in Wikipedia about the Hilbert-Polya conjecture that:

" ...a physical reason that the Riemann hypothesis should be true, and suggested that this would be the case if the imaginary parts $t$ of the zeros $$\frac{1}{2} + it$$ of the Riemann zeta function corresponded to eigenvalues of an unbounded self-adjoint operator."

What I have is this program that takes as input the imaginary parts of the tetrated non-trivial zeros of the Riemann zeta function and sums them over the divisors to form a matrix that has the Greatest Common Divisor matrix (GCD(n,k)) as lookup index. As an example of this matrix I give the case $5$ times $5$:

$$\begin{array}{ccccc} e^{e^{e^{e^{\frac{\Im(\rho _1)}{10 \pi }}}}} & e^{e^{e^{e^{\frac{\Im(\rho _1)}{10 \pi }}}}} & e^{e^{e^{e^{\frac{\Im(\rho _1)}{10 \pi }}}}} & e^{e^{e^{e^{\frac{\Im(\rho _1)}{10 \pi }}}}} & e^{e^{e^{e^{\frac{\Im(\rho _1)}{10 \pi }}}}} \\ e^{e^{e^{e^{\frac{\Im(\rho _1)}{10 \pi }}}}} & e^{e^{e^{e^{\frac{\Im(\rho _1)}{10 \pi }}}}}+e^{e^{e^{e^{\frac{\Im(\rho _2)}{10 \pi }}}}} & e^{e^{e^{e^{\frac{\Im(\rho _1)}{10 \pi }}}}} & e^{e^{e^{e^{\frac{\Im(\rho _1)}{10 \pi }}}}}+e^{e^{e^{e^{\frac{\Im(\rho _2)}{10 \pi }}}}} & e^{e^{e^{e^{\frac{\Im(\rho _1)}{10 \pi }}}}} \\ e^{e^{e^{e^{\frac{\Im(\rho _1)}{10 \pi }}}}} & e^{e^{e^{e^{\frac{\Im(\rho _1)}{10 \pi }}}}} & e^{e^{e^{e^{\frac{\Im(\rho _1)}{10 \pi }}}}}+e^{e^{e^{e^{\frac{\Im(\rho _3)}{10 \pi }}}}} & e^{e^{e^{e^{\frac{\Im(\rho _1)}{10 \pi }}}}} & e^{e^{e^{e^{\frac{\Im(\rho _1)}{10 \pi }}}}} \\ e^{e^{e^{e^{\frac{\Im(\rho _1)}{10 \pi }}}}} & e^{e^{e^{e^{\frac{\Im(\rho _1)}{10 \pi }}}}}+e^{e^{e^{e^{\frac{\Im(\rho _2)}{10 \pi }}}}} & e^{e^{e^{e^{\frac{\Im(\rho _1)}{10 \pi }}}}} & e^{e^{e^{e^{\frac{\Im(\rho _1)}{10 \pi }}}}}+e^{e^{e^{e^{\frac{\Im(\rho _2)}{10 \pi }}}}}+e^{e^{e^{e^{\frac{\Im(\rho _4)}{10 \pi }}}}} & e^{e^{e^{e^{\frac{\Im(\rho _1)}{10 \pi }}}}} \\ e^{e^{e^{e^{\frac{\Im(\rho _1)}{10 \pi }}}}} & e^{e^{e^{e^{\frac{\Im(\rho _1)}{10 \pi }}}}} & e^{e^{e^{e^{\frac{\Im(\rho _1)}{10 \pi }}}}} & e^{e^{e^{e^{\frac{\Im(\rho _1)}{10 \pi }}}}} & e^{e^{e^{e^{\frac{\Im(\rho _1)}{10 \pi }}}}}+e^{e^{e^{e^{\frac{\Im(\rho _5)}{10 \pi }}}}} \end{array}$$

In the example matrix above the number of exponentiations is finite, but they are supposed to be infinitely many as in tetration like this:

$$\underbrace{e^{e^{\cdot^{\cdot^{e^{\frac{\Im (\rho_1)}{10\pi}}}}}}}_n$$

where $n \rightarrow \infty$

Division by $10\pi$ is there to avoid overflow in the program. Notice also that it is a sum over the divisors and that there are plus $+$ signs in between some of the terms.

Anyways as a Mathematica program this:

Input:

14.1347251417,21.0220396388,25.0108575801,30.4248761259,32.9350615877

---

(*Mathematica 8 program start*)
Clear[a, A1, A2, a2, a1, A3, nn, n, k, d, kolumn]
a[n_] := If[n < 1, 0, Sum[Im[ZetaZero[d]]/(10*Pi), {d, Divisors[n]}]]
Do[nn = j;
 A3 = Range[nn]*0;
 Do[kolumn = i;
   A1 = Table[Table[a[GCD[n, k]], {k, 1, nn}], {n, 1, nn}];
   MatrixForm[A1];
   A1[[All, kolumn]];
   MatrixForm[
    Table[Table[
      If[Mod[n, k] == 0, MoebiusMu[n/k]*A1[[All, kolumn]][[k]], 
       0], {k, 1, nn}], {n, 1, nn}]];
   a1 = Table[
     Total[Table[
       If[Mod[n, k] == 0, MoebiusMu[n/k]*A1[[All, kolumn]][[k]], 
        0], {k, 1, nn}]], {n, 1, nn}];
   a2 = Sign[a1]*Exp[Exp[Exp[Exp[Abs[a1]]]]];
   A2 = Table[
     Table[If[Mod[n, k] == 0, a2[[n/k]], 0], {k, 1, nn}], {n, 1, nn}];
   MatrixForm[A2];
   a3 = Table[
     Total[Table[If[Mod[n, k] == 0, a2[[n/k]], 0], {k, 1, nn}]], {n, 
      1, nn}];
   A3[[i]] = a3;, {i, 1, nn}]
   MatrixForm[A3];
 Print[10*Pi*
   Reverse[N[
     Sign[Eigenvalues[A3]] Log[Log[Log[Log[Abs[Eigenvalues[A3]]]]]], 
     12]]], {j, 1, 5}]
MatrixForm[A3]
(*program end*)

with eigenvalues as output:

{14.1347251417}
{14.1347251417,21.0220396388}
{14.1347251417,21.0220396388,25.0108575801}
{14.1347251417,21.0220396388,25.0108575801,30.4248761259}
{14.1347251417,21.0220396388,25.0108575801,30.4248761259,32.9350615877}

Compared to the actual zeros:

N[Table[Im[ZetaZero[n]], {n, 1, 5}], 12]
{14.1347251417, 21.0220396388, 25.0108575801, 30.4248761259, 32.9350615877}

So oversimplified the program works like this:

My question then is:

Is the matrix I presented an unbounded self-adjoint operator?

I don't know what unbounded means other than that it has to do with this inequality from wikipedia about bounded operators:

$$\|Lv\|_Y \le M \|v\|_X.\, \,$$

I know that the identity matrix is Hermitian and that when it is element-wise multiplied with a vector of arbitrary numbers, it will have those same numbers as eigenvalues. Although I don't know why, the identity matrix is hardly an "unbounded self-adjoint operator". What I do know is that the matrix above in this question is related to another simpler matrix that satisfies similar recurrences as the identity matrix.

---