Let $A = (a_{ij})_{1\le i,j\le h}$ be an $h$-by-$h$ non-degenerate upper triangular matrix with entry $a_{11} = 1$. Let $\Phi = \{\alpha_1,\alpha_2,\ldots,\alpha_d\}\subseteq \{1,2,\ldots,h\}=I$ be an ordered set satisfying:
- $\alpha_1 = 1$,
- the subsequence $\alpha_1,\ldots,\alpha_s$ for some $2\le s\le d$ is the longest subsequence of elements in $\Phi$ for which any $1\le j\le s-1$ and any $\alpha_i\in \Phi\setminus\{\alpha_1,\ldots,\alpha_{s}\}$, $|\alpha_j - \alpha_{j+1}| < |\alpha_i - \alpha_{i+1}|$. If multiple such sequences of elements in $\Phi$ exist, choose the one such that the next subsequence $\alpha_s,\ldots,\alpha_{s+r}$ for some $s+1\le s+r\le d$ satisfies the same conditions with respect to elements in $\Phi\setminus \{\alpha_1,\ldots,\alpha_s\}$, and repeat the minimization as necessary.
Assume that $\Phi$ further satisfies the condition that there is no integer $0<i<h-1$ such that $\Phi + i = \{\alpha_1+i,\ldots,\alpha_d+i\} = \Phi$, where addition is performed modulo $h$.
Then is there a way to show that for any $1\le k\le d$ that for $1\le s\le h$ the determinant of all the $k$-by-$k$ matrices $(a_{ij})$ where $i = \alpha_i + s$ ranges over the first $k$ elements of $\Phi$ and $1\le j\le k$ (where again addition is performed modulo $h$) cannot simultaneously vanish?
This is evidently true for $k=1,2$, and it is a basic exercise to show this for $\{\alpha_1,\ldots,\alpha_d\} = \{1,2,\ldots,d\}$ by using induction and a cofactor expansion. I am interested in the other cases.
Note: $i$ is used to index along the rows and $j$ is used to index along the columns in the present convention.
