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}) $.

  • 2
    $\begingroup$ Looks good to me. $\endgroup$
    – Daniel S
    Jun 2 at 7:51
  • $\begingroup$ Thank's Daniel! $\endgroup$
    – P_Gate
    Jun 2 at 8:14
  • $\begingroup$ it is correct you can close the question $\endgroup$
    – Don Freecs
    Jun 3 at 23:33
  • 1
    $\begingroup$ Really the OP should factor out their proof from the question, and answer their own question with that answer. $\endgroup$
    – Mark
    Jun 5 at 21:07
  • $\begingroup$ Didn't you post this question on Math SE? $\endgroup$ Jun 6 at 6:51

1 Answer 1


Thanks for the comments confirming that my proof is correct. As suggested in the comments, I'm happy to answer the question myself.


$\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 $\mathbf{z} \in \Lambda^{\perp}(\mathbf{HA})$.


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