Question on proof ##\Lambda^{\perp}(AU) = U^{-1} \Lambda^{\perp}(A)##

[Peter_Newman]

Say we have as special lattice $\Lambda^{\perp}(A) = \left\{z \in \mathbf{Z^m} : Az = 0 \in \mathbf{Z_q^n}\right\}$. We define $U \in \mathbf{Z^{m \times m}}$ as an invertible matrix then I want to proof the following fact:
$$ \Lambda^{\perp}(AU) = U^{-1} \Lambda^{\perp}(A) $$
My idea:
Let $y \in \Lambda^{\perp}(A)$ that is $y \in Az = 0$, now $U^{-1}y = (U^{-1}Az = 0) \in U^{-1}\Lambda^{\perp}(A)$ and let $y' \in \Lambda^{\perp}(AU)$ that is $y' \in AUz = 0$, this implies $y \in y'$ which shows one direction.

I hope that this is not too simple thinking and therefore I am interested in your opinions.

 

[pasmith]

Your notation is confusing. You don't mean $y \in Az = 0$ etc.; you mean $y \in \{ z \i n\mathbb{Z}^m: Az = 0 \}$, but you can just write "Let y \in \Lambda^{\perp}(A). Then $Ay = 0$."

The central point is that if $Ay = 0$ then we can write $Ay = AUU^{-1}y$ so that $U^{-1} y \in \Lambda^{\perp}(AU)$; hence $$U^{-1}\Lambda^{\perp}(A) \subset \Lambda^{\perp}(AU).$$ But conersely, if $AUy = 0$ then $Uy \in \Lambda^{\perp}(A)$ so that $$U\Lambda^{\perp}(AU) \subset \Lambda^{\perp}(A).$$ But if $U$ is invertible then $$U(B) = C \Leftrightarrow B = U^{-1}(C)$$ for any subsets $B$ and $C$ of $\mathbb{Z}^m$. The result follows.

The requirement that a matrix have integer entries and have an inverse with integer entries is somewhat restrictive; the only ones which come to mind are $\pm I$ and matrices which permute the standard basis vectors.

 

[Peter_Newman]

Thanks for your great help @pasmith ! Yes my notation is a bit confusing and also kept myself from seeing the result directly.

In the second part, you could have done the following: $y \in \Lambda^{\perp}(AU)$ then $AUy = 0$, then $Uy \in \Lambda^{\perp}(A)$ which implies $y \in U^{-1}\Lambda^{\perp}(A)$ , right? The "advantage" would be that then $\Lambda^{\perp}(AU) = U^{-1} \Lambda^{\perp}(A)$ is directly recognizable, but this is more or less a rewriting.