Zero Sums in a $q$-Deformation Remain Zero for $q=1$

Question

Looking at previous question, I begun to think and came upon the following more solid question: Let $SU_q(N)$ be the usual quantized coordinated algebra of $SU(N)$. Now consider, for some fixed value of $q$, the element $$ v := \sum_{i=1}^k \lambda^i(q) v_i \in SU_q(N), $$ where each $\lambda^i(q)$ is some polynomial function in $q$, and each $v_i$ is a product of some subset of the generators of $SU_q(N)$. It has the obvious corresponding element in ${\cal O}(SU(N))$: $$ v' := \sum_{i=1}^k \lambda^i(1) v_i.$$ It seems to be for me obvious that if $v =0$, then we must necessarily to have $v' = 0$. However, I can't see how to prove this with care?

Edit: I should say transcendental $q$ to make this question non-trivial

Answer 1

I think so (unless i missed a subtility), at least for N=2 (and the argument is probably similar in general). Namely, recall that there's a basis of the vector space $SL_q(2)$ consisting of $\{a^nb^mc^r,b^mc^rd^s,m,r,s\in N_o, n\in N\}$. So you can, wlog, assume that the $v_i's$ are of this form in your expression in $SL_q(2)$, with coefficient that will be polynomials in $q$. If you assume that the expression $v(q)$ vanishes for generic $q$'s, then these polynomials must be identically 0 (because you assumed they are polynomials, not any function).