A matrix of real numbers is called totally positive if all its minors are non-negative. A well-known example is the Pascal matrix $(\binom{i}{j})$.


Is it true that the minors of the $q$-Pascal matrix $({\binom{i}{j}}\_q)$ are polynomials with non-negative coefficients?



Total positivity of $q$-Pascal matrix?
Yes. Consider the set $V$ of points with integer coordinates as vertices of a weighted directed acyclic graph. Namely, for any $(i,j)\in V$, the edge from $(i,j)$ to $(i+1,j)$ has weight 1, the edge from $(i,j)$ to $(i,j+1)$ has weight $q^{i}$. Then $\binom{n+k}{k}\_q$ is a weighted sum of paths from the origin to $(n,k)$, it follows by induction from the $q$-Pascal identity $\binom{n+k}{k}\_q=q^k\binom{n+k-1}{k}\_q+\binom{n+k-1}{k-1}\_q$. For integers $a,b$ and non-negative integers $n,k$, the weighted sum of paths from $(a,b)$ to $(a+n,b+k)$ equals $q^{an}\binom{n+k}{k}\_q$, this is seen from comparing with paths between the origin and $(n,k)$. 


Now consider the minor $\binom{u\_i}{v\_j}\_q$, $i,j=1,\dots,m$ of your matrix. Denote $A\_j=(-v\_j,v\_j)$, $B\_i=(0,u\_i)$. Then the weighted sum of paths from $A\_j$ to $B\_i$ equals $q^{-v\_j^2}\binom{u\_i}{v\_j}\_q$. Thus our determinant is a Laurent polynomial with non-negative coefficients by [Lindström–Gessel–Viennot lemma](https://en.wikipedia.org/wiki/Lindstr%C3%B6m%E2%80%93Gessel%E2%80%93Viennot_lemma). On the other hand, the value of a minor is a genuine polynomial in $q$.


