update notations

content/CSE5313/CSE5313_L10.md

@@ -40,20 +40,20 @@ Let $G$ and $H$ be the generator and parity-check matrices of (any) linear code
 #### Lemma 1
 
 $$
-H G^T = 0
+H G^\top = 0
 $$
 
 <details>
 <summary>Proof</summary>
 
-By definition of generator matrix and parity-check matrix, $forall e_i\in H$, $e_iG^T=0$.
+By definition of generator matrix and parity-check matrix, $forall e_i\in H$, $e_iG^\top=0$.
 
-So $H G^T = 0$.
+So $H G^\top = 0$.
 </details>
 
 #### Lemma 2
 
-Any matrix $M\in \mathbb{F}_q^{(n-k)\times n}$ such that $\operatorname{rank}(M) = n - k$ and $M G^T = 0$ is a parity-check matrix for $C$ (i.e. $C = \ker M$).
+Any matrix $M\in \mathbb{F}_q^{(n-k)\times n}$ such that $\operatorname{rank}(M) = n - k$ and $M G^\top = 0$ is a parity-check matrix for $C$ (i.e. $C = \ker M$).
 
 <details>
 <summary>Proof</summary>
@@ -62,7 +62,7 @@ It is sufficient to show that the two statements
 
 1. $\forall c\in C, c=uG, u\in \mathbb{F}^k$
 
-$M c^T = M(uG)^T = M(G^T u^T) = 0$ since $M G^T = 0$.
+$M c^\top = M(uG)^\top = M(G^\top u^\top) = 0$ since $M G^\top = 0$.
 
 Thus $C \subseteq \ker M$.
 
@@ -84,15 +84,15 @@ We proceed by applying the lemma 2.
 
 1. $\operatorname{rank}(H) = n - k$ since $H$ is a Vandermonde matrix times a diagonal matrix with no zero entries, so $H$ is invertible.
 
-2. $H G^T = 0$.
+2. $H G^\top = 0$.
 
-note that $\forall$ row $i$ of $H$, $0\leq i\leq n-k-1$, $\forall$ column $j$ of $G^T$, $0\leq j\leq k-1$
+note that $\forall$ row $i$ of $H$, $0\leq i\leq n-k-1$, $\forall$ column $j$ of $G^\top$, $0\leq j\leq k-1$
 
 So
 
 $$
 \begin{aligned}
-H G^T &= \begin{bmatrix}
+H G^\top &= \begin{bmatrix}
 1 & 1 & \cdots & 1\\
 \alpha_1 & \alpha_2 & \cdots & \alpha_n\\
 \alpha_1^2 & \alpha_2^2 & \cdots & \alpha_n^2\\