We have the following result: Let $n\in\mathbb{N}_ {\ge 3}$ and $f_k\in(0, +\infty)$ be such that $f_{k+n}=f_{k}$, for all $k\in\mathbb{Z}$. Furthermore, let $$\gamma_k=-\sqrt{\frac{f_{k-1}f_{k+1}}{(f_{k-1}+f_k)(f_k+f_{k+1})}}$$ and $M$ be the $n\times n$ matrix such that $M=(m_{ij})$, where $$m_{ij}=\begin{cases} 1 & \text{if } i=j; \\\\ \gamma_i & \text{if } i\equiv j+1 \pmod{n}; \\\\ \gamma_j & \text{if } j\equiv i+1\pmod{n}; \\\\ 0 & \text{ otherwise}. \end{cases}$$ Then $M$ is singular and its kernel is generated by a vector $v\in(0,+\infty) ^{n}$.
Although we were able to prove this, we relied on combinatorial arguments like the inclusion-exclusion principle (see the outline of the proof below), and this kinda left a "bitter aftertaste" since we were sure that this matrix should have some nice algebraic properties, and we were expecting an algebraic argument to emerge naturally, but our efforts in that direction were unsuccessful. Is there potential for an algebraic argument, perhaps involving a factorization of $M$, that we have overlooked?
Here is an overview of the proof: It is possible to prove that $$\det M=1+2(-1)^{n-1}\gamma_1\dots\gamma_n+\sum_{k=1}^{n}(-1)^k\sum_{[i_1, \dots, i_k]}\gamma_{i_1}^2\dots\gamma_{i_k}^2,$$ where the symbol $[i_1, \dots, i_k]$ means that we take the summation over all the k-tuples $(i_1, \dots, i_k)$ such that $1\le i_1<\dots<i_k\le n$ and $i_j, i_{j+1}$ are not consecutive (we consider $n$ and $1$ to be consecutive). Set $a_k=f_{k-1}/(f_{k-1}+f_k)$, so that $\gamma_k^2=a_k(1-a_{k+1})$. Let $(\Omega, P)$ be a probability space, $\\\{ A_k \} _{1\le k\le n}$ be mutually independent events of $\Omega$ such that $P(A_k)=a_k$ and $E_k=A_k\cap A _{k+1} ^c$. Using a double counting type of argument on $P(E_1\cup\dots\cup E_n)$, one can show that $\det M=0$. To find the kernel of $M$, one can begin by doing a brute-force row reduction on the matrix $M=0$ and use again the events $E_i$ to show that if $v=(x_1, \dots, x_n)$ is a non-trivial solution to $Mv=0$ such that $x _{n-1}>0$, then $x_n>0$, and the result follows from symetry arguments.