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 [itex]y \in Az = 0[/itex] etc.; you mean [itex]y \in \{ z \i n\mathbb{Z}^m: Az = 0 \}[/itex], but you can just write "Let y \in \Lambda^{\perp}(A). Then [itex]Ay = 0[/itex]."

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