How to prove the inequalities of q-ary lattice determinant?

for $A\in{Z_q^{n*m}}$ and $A^{'}\in{Z_q^{m*n}}$,we have

• $det{({\land}_q^{\bot}(A))}{\le}q^n$ and $det{({\land}_q(A^{'}))}{\ge}q^{m-n}$
• if q is prime,and A,A' are non-singular in the finite field $Z_q$,the above inequality are equalities.

where ${\land}_q^{\bot}(A) = \{x{\in}Z^m|Ax=0{\bmod}q\}$ and ${\land}_q(A)=\{y{\in}Z^m|y=As{\bmod}q\}$

The above content comes from D. Dadush's lecture note(lemma 4) lecture_9.I don't know how to prove the above lemma because the proof in the lecture note is too sketchy for me.I would appreciate it if someone could provide more detailed proof。

I get inspiration from vadim's ppt.But I only prove half of the theorem.

Proof: ${\because}{\land_q^{\bot}}(A)$ is a integer lattice, ${\therefore} {det({\land_q^{\bot}}(A))}=|{Z^m}/{\land_q^{\bot}}(A)|$. let us define a mapping ${f}:{Z^m}{\to}{Z_q^{n}}$,${f:Ax{\bmod}q}$.It's easy to verify that $f$ is homomorphic.According to the basic theorem of homomorphism，$|{Z^m}/kerf|=|{Z^m}/{\land_q^{\bot}}(A)|=|im({Z^m})|$.Because $im({Z^m}){\subseteq}{Z_q^{n}}$,so ${det({\land_q^{\bot}}(A))}=|{Z^m}/{\land_q^{\bot}}(A)|{\leq}q^n$. If q is prime,and A is non-singular,then $f$ is full homomorphism because every image in ${Z_q^{n}}$ can find the original image in ${Z^m}$.So $im({Z^m})={Z_q^{m}}$ and ${det({\land_q^{\bot}}(A))}=|{Z^m}/{\land_q^{\bot}}(A)|=q^n$.

However,I haven't found a way to prove the other half of the theorem. I would be grateful if someone could give the rest of the proof.