Is there a determinant that in the limit becomes the Riemann zeta function?

Question

Let the matrix A be: $$A(n,k): \text{If } n=k \; \text{ then } 1+\frac{1}{n^{\frac{1}{4} (3+2 i t)} k^{\frac{1}{4} (3+2 i t)}}\text{ else }\frac{1}{n^{\frac{1}{4} (3+2 i t)} k^{\frac{1}{4} (3+2 i t)}}$$

which is equivalent to: $$A(n,k): \text{If } n=k \; \text{ then } 1+\frac{1}{n^{\frac{1}{2} (3+2 i t)}}\text{ else }\frac{1}{n^{\frac{1}{4} (3+2 i t)} k^{\frac{1}{4} (3+2 i t)}}$$

which is equivalent to $A=M+I$ where $M$ equals:

$M(n,k)=\frac{1}{n^{\frac{1}{4} (3+2 i t)} k^{\frac{1}{4} (3+2 i t)}}$

and $I$ is the identity matrix in linear algebra:

$$I(n,k): \text{If } n=k \; \text{ then } 1 \text{ else } 0$$

The first few determinants starting at the upper left corner of the infinite matrix $A$ are:

The $1$x$1$ matrix: $$2$$ The $2$x$2$ matrix: $$2^{1+\frac{1}{2} (-3-2 i t)}-2^{\frac{1}{2} (-3-2 i t)}+2$$ The $3$x$3$ matrix: $$2^{1+\frac{1}{2} (-3-2 i t)}-2^{\frac{1}{2} (-3-2 i t)}+3^{\frac{1}{2} (-3-2 i t)}+2$$ ...

The plot of the determinant of the $10$ x $10$ matrix looks like this:

where the vertical lines are at the Riemann zeta zeros on the critical line.

Compare this to the actual Riemann zeta function on the critical line:

Is there a infinite sized matrix, which is not totally degenerate, such that the determinant is the Riemann zeta function?

Associated Mathematica program:

(*start*)
Clear[n, k, nn, t, x];
nn = 10;
A = Table[
   Table[If[n == k, 1 + 1/n^(1/4 (3 + 2 I t))/k^(1/4 (3 + 2 I t)), 
     1/n^(1/4 (3 + 2 I t))/k^(1/4 (3 + 2 I t))], {k, 1, nn}], {n, 1, 
    nn}];
pol = Det[A];
Show[Plot[Re[pol], {t, 0, 60}],
 Graphics[
  Table[Line[{{Im[ZetaZero[n]], -30}, {Im[ZetaZero[n]], 30}}], {n, 1, 
    12}]], ImageSize -> Large]
Plot[Re[Zeta[1/2 + I*t]], {t, 0, 60}, ImageSize -> Large]
(*end*)

Answer 1

Mathematica:

Table[M = 
  Table[Table[
    If[Mod[n, k] == 0, MoebiusMu[n/k]/(n/k)^s, If[n == 1, 1, 0]], {k, 
     1, nn}], {n, 1, nn}];
 Expand[FullSimplify[Det[M]]], {nn, 1, 12}]

Output:

{1, 1 + 2^-s, 1 + 2^-s + 3^-s, 1 + 2^-s + 3^-s + 4^-s, 1 + 2^-s + 3^-s + 4^-s + 5^-s, 1 + 2^-s + 3^-s + 4^-s + 5^-s + 6^-s, 1 + 2^-s + 3^-s + 4^-s + 5^-s + 6^-s + 7^-s, 1 + 2^-s + 3^-s + 4^-s + 5^-s + 6^-s + 7^-s + 8^-s, 1 + 2^-s + 3^-s + 4^-s + 5^-s + 6^-s + 7^-s + 8^-s + 9^-s, 1 + 2^-s + 3^-s + 4^-s + 5^-s + 6^-s + 7^-s + 8^-s + 9^-s + 10^-s, 1 + 2^-s + 3^-s + 4^-s + 5^-s + 6^-s + 7^-s + 8^-s + 9^-s + 10^-s + 11^-s, 1 + 2^-s + 3^-s + 4^-s + 5^-s + 6^-s + 7^-s + 8^-s + 9^-s + 10^-s + 11^-s + 12^-s}

https://oeis.org/A143142

Answer 2

I construct the matrix, for instance size 4x4, symbolically, giving the entries at r'th row and c'th column symbolical names which we shall later expand into the actual formulae: $$ A_4= \small \begin{bmatrix} c_1 r_1+1 & c_2 r_1 & c_3 r_1 & c_4 r_1 \\ r_2 c_1 & c_2 r_2+1 & c_3 r_2 & c_4 r_2 \\ r_3 c_1 & c_2 r_3 & c_3 r_3+1 & c_4 r_3 \\ r_4 c_1 & c_2 r_4 & c_3 r_4 & c_4 r_4+1 \end{bmatrix} \tag 1 $$ The determinant of this matrix is $$ det(A_4) = 1+ c_1 r_1 + c_2 r_2 + c_3 r_3 + c_4 r_4 \tag 2$$ So this is a very simple expression and comparing with larger matrix-sizes we find the obvious generalization.
Now we replace the symbolic names by the actually intended terms $$r_k=c_k = k^{-1/4(3+2it)}$$ and have the determinant $$ \det(A_4) =1+ 1 \cdot 1 + 2^{{-1/4(3+2it)}} \cdot 2^{-1/4(3+2it)} + 3^{{-1/4(3+2it)}} \cdot 3^{-1/4(3+2it)} + 4^{{-1/4(3+2it)}} \cdot 4^{-1/4(3+2it)} \\ =2 + 2^{-1/2(3+2it)} + 3^{-1/2(3+2it)} + 4^{-1/2(3+2it)} $$ The function to plot, in additional dependance of the matrixsize $n$ is the real part of $$f_n(t) = 2 + \sum_{k=2}^n {1\over k^{ \, 3/2 + \, î \, t}} \tag 3$$ which is simply the partial-sum evaluation of the zeta at $s=3/2+ î t$ and can thus be written without all the matrix-overhead.

I wouldn't see, why the determinant-expression (2) should not be extended formally to infinite matrices as long it designs a convergent sum, or an expression which could be analytically continued for the infinite-size case.

(Remark: I do not yet understand why we have the real part of the exponent being 3/2 instead of 1/2, is this perhaps a typo in the question?)

Answer 3

Fix $N,s$. Your $N \times N$ matrix is $A = I+u^Tu$

where $u = (1,2^{-s},3^{-s},\ldots,N^{-s})$ is a row vector.

Hence thanks to the determinant lemma $$\det(A) = (1+u u^T)\det(I) = 1+\sum_{n=1}^N n^{-2s}$$

and Sherman-Morrison gives $B = A^{-1} = I - \frac{ u^Tu}{1+uu^T}$ i.e. $B(n,k)= 1_{n=k}-\frac{(nk)^{-s}}{\det(A)}$