# Proof regarding a property of "$q$-ary" lattices

In this question we are dealing with "$$q$$-ary" lattices. I will give the definition available to me and I'm interested in proving the lemma. As a reference see the PDF on page 2 from Peikert's lectures.

Definition. Let $$\mathbb{Z}_q := \{ 0, 1, \dots, q-1\}$$. We define $$\Lambda^{\perp}(\mathbf{A}) := \left\{ \mathbf{z} \in \mathbb{Z^m} : \mathbf{Az = 0} \right\}$$, where $$\mathbf{A} \in \mathbb{Z}_q^{n \times m}$$

Lemma. Let $$\mathbf{H} \in \mathbb{Z}_{q}^{n \times n}$$ be invertible. Then $$\Lambda^{\perp}(\mathbf{HA}) = \Lambda^{\perp}(\mathbf{A})$$.

• Looks good to me. Jun 2, 2023 at 7:51
• Thank's Daniel! Jun 2, 2023 at 8:14
• it is correct you can close the question Jun 3, 2023 at 23:33
• Really the OP should factor out their proof from the question, and answer their own question with that answer. Jun 5, 2023 at 21:07
• Didn't you post this question on Math SE? Jun 6, 2023 at 6:51

$$\Lambda^{\perp}(\mathbf{HA}) \subset \Lambda^{\perp}(\mathbf{A})$$, let $$\mathbf{z} \in \Lambda^{\perp}(\mathbf{HA})$$ then $$\mathbf{HAz = 0}$$ and $$H$$ is invertible implies $$\mathbf{H^{-1}HAz = 0}$$ and this implies $$\mathbf{z} \in \Lambda^{\perp}(\mathbf{A})$$.
$$\Lambda^{\perp}(\mathbf{A}) \subset \Lambda^{\perp}(\mathbf{HA})$$, let $$\mathbf{z} \in \Lambda^{\perp}(\mathbf{A})$$ then $$\mathbf{Az = 0}$$ and multiply by $$\mathbf{H}$$ implies $$\mathbf{HAz = 0}$$ and this implies $$\mathbf{z} \in \Lambda^{\perp}(\mathbf{HA})$$.