Let $f(n)$ be a complex sequence. Then for prime $p$ define $\hat{f}(p) = \sum_{n = 1}^{\infty} a_n e^{-i 2 \pi n / p}$. Then let the transformation of sequences be $T$, i.e. $Tf = \hat{f}$. Is there an inverse transform? The transformation result sequence is complex, indexed by the primes.
In other words, does the infinite matrix $$ \begin{pmatrix} z^{k/p_j} & \cdots \\ \vdots & \ddots \end{pmatrix} \text{ where } z = e^{-i 2 \pi} \text{ and element $z^{k/p_j}$ is in row $j$, column $k$, $p_j$ is the $j$th prime.} $$ have an inverse.
How would you solve this?
Let's relax the restriction that $\hat{f}$ sequence must be prime indexed and say it has the same index as $f$.
Let's see if finite dimensioned square matrices of this form have an inverse.
Let $A = $
$$ \begin{pmatrix} z^{1/2} & z \\ z^{1/3} & z^{2/3} \end{pmatrix} $$
It's $\det$ is $z^{1/2}z^{2/3} - zz^{1/3} = z^{7/6} - z^{4/3} \neq 0$ since $z^{7/6 - 4/3 = -1/6} \neq 1$.
It's inverse is $\cdots$
This seems tedious. A computer should do it.