update notations

content/Math429/Math429_L12.md

@@ -25,7 +25,7 @@ Let $A$ be an $m \times n$ matrix, then
 * The column rank of $A$ is the dimension of the span of the columns in $\mathbb{F}^{m,1}$.
 * The row range of $A$ is the dimension of the span of the row in $\mathbb{F}^{1,n}$.
 
-> Transpose: $A^t=A^T$ refers to swapping rows and columns
+> Transpose: $A^t=A^\top$ refers to swapping rows and columns
 
 #### Theorem 3.56 (Column-Row Factorization)
 
@@ -64,7 +64,7 @@ Proof:
 
 Note that by **Theorem 3.56**, if $A$ is $m\times n$ and has column rank $c$. $A=CR$ for some $C$ is a $m\times c$ matrix, $R$ is a $c\times n$ matrices, ut the rows of $CR$ are a linear combination of the rows of $R$, and row rank of $R\leq C$. So row rank $A\leq$ column rank of $A$.
 
-Taking a transpose of matrix, then row rank of $A^T$ (column rank of $A$) $\leq$ column rank of $A^T$ (row rank $A$).
+Taking a transpose of matrix, then row rank of $A^\top$ (column rank of $A$) $\leq$ column rank of $A^\top$ (row rank $A$).
 
 So column rank is equal to row rank.
 

content/Math429/Math429_L18.md

@@ -39,13 +39,13 @@ $T$ is surjective $\iff range\ T=W\iff null\ T'=0\iff T'$ injective
 
 Let $V,W$ be a finite dimensional vector space, $T\in \mathscr{L}(V,W)$
 
-Then $M(T')=(M(T))^T$. Where the basis for $M(T)'$ are the dual basis to the ones for $M(T)$
+Then $M(T')=(M(T))^\top$. Where the basis for $M(T)'$ are the dual basis to the ones for $M(T)$
 
 #### Theorem 3.133
 
 $col\ rank\ A=row\ rank\ A$
 
-Proof: $col\ rank\ A=col\ rank\ (M(T))=dim\ range\ T=dim\ range\ T'=dim\ range\ T'=col\ rank\ (M(T'))=col\ rank\ (M(T)^T)=row\ rank\ (M(T))$
+Proof: $col\ rank\ A=col\ rank\ (M(T))=dim\ range\ T=dim\ range\ T'=dim\ range\ T'=col\ rank\ (M(T'))=col\ rank\ (M(T)^\top)=row\ rank\ (M(T))$
 
 ## Chapter V Eigenvalue and Eigenvectors