From b9cc74650eede4819bfd5f730b6039687ae23d02 Mon Sep 17 00:00:00 2001
From: Sun Yimin <emmansun@users.noreply.github.com>
Date: Wed, 16 Apr 2025 01:54:43 +0000
Subject: [PATCH] =?UTF-8?q?Updated=20=E5=90=8E=E9=87=8F=E5=AD=90=E5=AF=86?=
 =?UTF-8?q?=E7=A0=81=E5=AD=A6=EF=BC=88PQC=EF=BC=89=E2=80=90=20=E5=AE=9E?=
 =?UTF-8?q?=E7=8E=B0Kyber=E6=89=80=E9=9C=80=E7=9A=84=E5=A4=9A=E9=A1=B9?=
 =?UTF-8?q?=E5=BC=8F=E5=92=8C=E7=BA=BF=E6=80=A7=E4=BB=A3=E6=95=B0=E7=9F=A5?=
 =?UTF-8?q?=E8=AF=86=20(markdown)?=
MIME-Version: 1.0
Content-Type: text/plain; charset=UTF-8
Content-Transfer-Encoding: 8bit

---
 ...r所需的多项式和线性代数知识.md | 18 +++++++++---------
 1 file changed, 9 insertions(+), 9 deletions(-)

diff --git a/后量子密码学（PQC）‐-实现Kyber所需的多项式和线性代数知识.md b/后量子密码学（PQC）‐-实现Kyber所需的多项式和线性代数知识.md
index e02b137..e42522f 100644
--- a/后量子密码学（PQC）‐-实现Kyber所需的多项式和线性代数知识.md
+++ b/后量子密码学（PQC）‐-实现Kyber所需的多项式和线性代数知识.md
@@ -94,22 +94,22 @@ $$A=\begin{bmatrix}
 
 ```math
 \hat{A} \cdot \hat{v} = 
-\begin{pmatrix}
+\begin{bmatrix}
     \hat{a}_{0,0} & \hat{a}_{0,1} & \hat{a}_{0,2} \\
     \hat{a}_{1,0} & \hat{a}_{1,1} & \hat{a}_{1,2} \\
     \hat{a}_{2,0} & \hat{a}_{2,1} & \hat{a}_{2,2} \\
-\end{pmatrix}
-\begin{pmatrix}
+\end{bmatrix}
+\begin{bmatrix}
     \hat{v}_{0} \\
     \hat{v}_{1} \\
     \hat{v}_{2} \\
-\end{pmatrix}
+\end{bmatrix}
 =
-\begin{pmatrix}
+\begin{bmatrix}
     \hat{a}_{0,0}\hat{v}_{0} + \hat{a}_{0,1}\hat{v}_{1} + \hat{a}_{0,2}\hat{v}_{2} \\
     \hat{a}_{1,0}\hat{v}_{0} + \hat{a}_{1,1}\hat{v}_{1} + \hat{a}_{1,2}\hat{v}_{2} \\
     \hat{a}_{2,0}\hat{v}_{0} + \hat{a}_{2,1}\hat{v}_{1} + \hat{a}_{2,2}\hat{v}_{2} \\
-\end{pmatrix}
+\end{bmatrix}
 ```
 
 当我们将一个向量（记为 $\hat{v}$ ）进行转置后与它自己相乘（即 $\hat{v}^T \circ \hat{v}$ ，这也被称为内积），其结果是一个 $1×k$ 维的矩阵与一个 $k×1$ 维的矩阵相乘，最终得到一个 $1×1$ 的矩阵。实际上，这只不过是说它是一个点积的复杂表达方式，其结果是产生一个“标量”（即一个单独的元素）。你只需要按坐标逐个相乘这些元素，然后将它们全部加在一起。
@@ -118,14 +118,14 @@ $$A=\begin{bmatrix}
 
 ```math
 \hat{u}^T \cdot \hat{v} = 
-\begin{pmatrix}
+\begin{bmatrix}
     \hat{u}_0 & \hat{u}_1 & \hat{u}_2 \\
-\end{pmatrix}
+\end{bmatrix}
 \begin{pmatrix}
     \hat{v}_{0} \\
     \hat{v}_{1} \\
     \hat{v}_{2} \\
-\end{pmatrix}
+\end{bmatrix}
 =\hat{u}_0\hat{v}_{0} + \hat{u}_1\hat{v}_{1} + \hat{u}_2\hat{v}_{2}
 
 ```