Total positivity of $q$-Pascal matrix? 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?
 A: 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. On the other hand, the value of a minor is a genuine polynomial in $q$.
