Does knowing the associated determinant show that the Hilbert-Polya conjecture is meaningless, or is this approach too naive?

Question

Let $A$ be the matrix:

$$A = \text{If } n \bmod k=0 \text{ then } 1\text{ else }0$$

Let $B$ be the matrix:

$$B=\text{If }k \bmod n=0\text{ then }\Im(\rho _n)\text{ else }0$$

where $\Im(\rho _n)$ is the imaginary part of the $n$-th Riemann zeta zero.

Consider the equation:

$$|A.B-A x|=0$$

which is almost as the eigenvalue equation (where || is the symbol for the determinant). The Hilbert-Polya conjecture is that the imaginary part of the Riemann zeta zeros are the eigenvalues of some unbounded self-adjoint operator.

I am not really sure what an unbounded self-adjoint operators is, but I have understood that a Hermitian matrix will do. Or in this case the infinite symmetric matrix $A.B$ starting:

$$A.B=\left( \begin{array}{ccccccc} 14.1347 & 14.1347 & 14.1347 & 14.1347 & 14.1347 & 14.1347 & \dots\\ 14.1347 & 35.1568 & 14.1347 & 35.1568 & 14.1347 & 35.1568 \\ 14.1347 & 14.1347 & 39.1456 & 14.1347 & 14.1347 & 39.1456 \\ 14.1347 & 35.1568 & 14.1347 & 65.5816 & 14.1347 & 35.1568 \\ 14.1347 & 14.1347 & 14.1347 & 14.1347 & 47.0698 & 14.1347 \\ 14.1347 & 35.1568 & 39.1456 & 35.1568 & 14.1347 & 97.7538 \\ \vdots & & & & & & \ddots \end{array} \right)$$

The solutions to the equation:

$$|A.B-A x|=0$$

are exactly the imaginary parts of the Riemann zeta zeros, as shown by this Mathematica program:

Clear[x, polynomial, A, B];
TableForm[polynomial = Table[
   A = Table[Table[If[Mod[n, k] == 0, 1, 0], {k, 1, nn}], {n, 1, nn}];
   B = Table[
     Table[If[Mod[k, n] == 0, Im[ZetaZero[n]], 0], {k, 1, nn}], {n, 1,
       nn}];
   x /. Solve[Det[A.B - x*A] == 0, x], {nn, 1, 6}]]
MatrixForm[N[A.B]];

with the output:

{14.1347, 21.022, 25.0109, 30.4249, 32.9351, 37.5862}

But that is not very surprising because we have merely stated a way of doing Möbius inversion by finding the roots of the mentioned polynomial.

If you still are interested in zeta zeros as eigenvalues, then I say that it is the same matrix that comes closest, only the equation to solve becomes:

$$|A.B-xI|=0$$ or: $$|A.B-\lambda I|=0$$ with $\lambda$ instead of $x$, as more commonly seen.

My question is: What is meant by the unbounded self-adjoint operator in the Hilbert-Polya conjecture if not the associated determinant to the polynomials $p_1$ to $p_k$?:

$$\begin{array}{l} p_1 = \Im(\rho _1)-x \\ p_2 = (x-\Im(\rho _1)) (x-\Im(\rho _2)) \\ p_3 = -(x-\Im(\rho _1)) (x-\Im(\rho _2)) (x-\Im(\rho _3)) \\ p_4 = (x-\Im(\rho _1)) (x-\Im(\rho _2)) (x-\Im(\rho _3)) (x-\Im(\rho _4)) \\ p_5 = -(x-\Im(\rho _1)) (x-\Im(\rho _2)) (x-\Im(\rho _3)) (x-\Im(\rho _4)) (x-\Im(\rho _5)) \\ p_6 = (x-\Im(\rho _1)) (x-\Im(\rho _2)) (x-\Im(\rho _3)) (x-\Im(\rho _4)) (x-\Im(\rho _5)) (x-\Im(\rho _6)) \end{array}$$

with the simple definition:

$$p_k = (-1)^k \prod _{n=1}^k (x-\Im(\rho _n))$$

There is a factor $(-1)^k$ which I don't know where it comes from other than that it comes from this program:

Clear[x, polynomial, A, B];
TableForm[polynomial = Table[
   A = Table[Table[If[Mod[n, k] == 0, 1, 0], {k, 1, nn}], {n, 1, nn}];
   B = Table[
     Table[If[Mod[k, n] == 0, Im[ZetaZero[n]], 0], {k, 1, nn}], {n, 1,
       nn}];
   Factor[Det[A.B - x*A]], {nn, 1, 6}]]

The zeros to the polynomials are the same, the Riemann zeta zeros, anyways.

I and others too can think of other determinants that are directly related to Riemann zeta zeros as eigenvalues, but they appear to be both tautological and degenerate.

This matrix $A.B$ is also tautological.

Also, can there be better determinants than the one shown here, and is this matrix $A.B$ degenerate?