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$$(\wedge_q(A))^\ast=\{y\in R^m: (y,z)\in Z \text{for any} z\in \wedge_q(A)\}.$$ If given $x\in(\wedge_q(A))^\ast$, how to prove $qx\in Z^m$?

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

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$$(\wedge_q(A))^\ast=\{y\in R^m: (y,z)\in Z \text{for any} z\in \wedge_q(A)\}.$$ If given $x\in(\wedge_q(A))^\ast$, how to prove $qx\in Z^m$?

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$$(\wedge_q(A))^\ast=\{y\in R^m: (y,z)\in Z \text{for any} z\in \wedge_q(A)\}.$$ If given $x\in(\wedge_q(A))^\ast$, how to prove $qx\in Z^m$?