For any matrix $A \in Z^{n\times m}$, Let $$\wedge_q(A)={ y\in Z^m:\exists s\in Z^n, s.t. y=A^ts (mod q) }$$, $$\wedge_q^\bot(A)=\{x\in z^m: Ax=0 mod q\}$$. There is a result stating that $q(\wedge_q(A))^\ast=\wedge_q^\bot(A)$, where$(\wedge_q(A))^\ast=\{y\in R^m: (y,z)\in Z for any z\in \wedge_q(A)\}$. If given $x\in(\wedge_q(A))^\ast$, how to prove $qx\in Z^m$?