norm symbol '|' to '||', R, C fonts

kf-liu · Sep 30, 2022 · 412afe9 · 412afe9
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diff --git a/The-Art-of-Linear-Algebra-j.pdf b/The-Art-of-Linear-Algebra-j.pdf
diff --git a/The-Art-of-Linear-Algebra-j.tex b/The-Art-of-Linear-Algebra-j.tex
@@ -268,6 +268,8 @@ \section{実用的なパターン}
 
 行列の分解については次の節でより詳しく解説する．
 
+\clearpage
+
 \section{5つの行列分解}
 
 \begin{itemize}
@@ -424,13 +426,12 @@ \subsection{$\boldsymbol{A=QR}$}
 このプロセスを順に続けていく．
 
 \begin{align*}
-  \bm{q}_1 &= \bm{a}_1/|\bm{a}_1| \\
-  \bm{q}_2 &= \bm{a}_2 - (\bm{q}_1\transp \bm{a}_2)\bm{q}_1 , \quad \bm{q}_2 = \bm{q}_2/|\bm{q}_2| \\
-  \bm{q}_3 &= \bm{a}_3 - (\bm{q}_1\transp \bm{a}_3)\bm{q}_1 - (\bm{q}_2\transp \bm{a}_3)\bm{q}_2, \quad \bm{q}_3 = \bm{q}_3/|\bm{q}_3|
+  \bm{q}_1 &= \bm{a}_1/||\bm{a}_1|| \\
+  \bm{q}_2 &= \bm{a}_2 - (\bm{q}_1\transp \bm{a}_2)\bm{q}_1 , \quad \bm{q}_2 = \bm{q}_2/||\bm{q}_2|| \\
+  \bm{q}_3 &= \bm{a}_3 - (\bm{q}_1\transp \bm{a}_3)\bm{q}_1 - (\bm{q}_2\transp \bm{a}_3)\bm{q}_2, \quad \bm{q}_3 = \bm{q}_3/||\bm{q}_3||
 \end{align*}
 
-あるいは，$\bm{a}$ を左辺に移動すると，以下のように記述できる．
-
+あるいは，$\bm{a}$ を左辺に移動して $r_{ij} = \bm{q}_i\transp \bm{a}_j$ とすると，以下のように記述できる．
 \begin{align*}
   \bm{a}_1 &= r_{11}\bm{q}_1\\
   \bm{a}_2 &= r_{12}\bm{q}_1 + r_{22} \bm{q}_2\\
@@ -454,18 +455,13 @@ \subsection{$\boldsymbol{A=QR}$}
   \\
   Q Q\transp=Q\transp Q = I
 \end{gather*}
-
 \begin{figure}[H]
   \includegraphics[keepaspectratio, width=\linewidth]{QR-j.eps}
   \caption{$A=QR$}
 \end{figure}
-
-
 $A$ の列ベクトルは，正規直行化された $Q$ の列ベクトルへと変換される．
 $A$ の列ベクトルを再生するには，$Q$ と $R$ の掛け算を考えれば簡単である．
 ここで，パターン1 (P1) を再度見て視覚的に理解して欲しい．
-
-
 \clearpage
 
 \subsection{$\boldsymbol{S=Q \Lambda Q\transp}$}

diff --git a/The-Art-of-Linear-Algebra.pdf b/The-Art-of-Linear-Algebra.pdf
diff --git a/The-Art-of-Linear-Algebra.tex b/The-Art-of-Linear-Algebra.tex
@@ -436,12 +436,12 @@ \subsection{$\boldsymbol{A=QR}$}
 procedure goes on.
 
 \begin{align*}
-  \bm{q}_1 &= \bm{a}_1/|\bm{a}_1| \\
-  \bm{q}_2 &= \bm{a}_2 - (\bm{q}_1\transp \bm{a}_2)\bm{q}_1 , \quad \bm{q}_2 = \bm{q}_2/|\bm{q}_2| \\
-  \bm{q}_3 &= \bm{a}_3 - (\bm{q}_1\transp \bm{a}_3)\bm{q}_1 - (\bm{q}_2\transp \bm{a}_3)\bm{q}_2, \quad \bm{q}_3 = \bm{q}_3/|\bm{q}_3|
+  \bm{q}_1 &= \bm{a}_1/||\bm{a}_1|| \\
+  \bm{q}_2 &= \bm{a}_2 - (\bm{q}_1\transp \bm{a}_2)\bm{q}_1 , \quad \bm{q}_2 = \bm{q}_2/||\bm{q}_2|| \\
+  \bm{q}_3 &= \bm{a}_3 - (\bm{q}_1\transp \bm{a}_3)\bm{q}_1 - (\bm{q}_2\transp \bm{a}_3)\bm{q}_2, \quad \bm{q}_3 = \bm{q}_3/||\bm{q}_3||
 \end{align*}
 
-or you can write:
+or you can write with $r_{ij} = \bm{q}_i\transp \bm{a}_j$:
 
 \begin{align*}
   \bm{a}_1 &= r_{11}\bm{q}_1\\