京都大学 情報学研究科 知能情報学専攻 2022年8月実施 情報学基礎 F1-1 fix

docs/kakomonn/kyoto_university/informatics/ist_202208_kiso_f1_1.md

@@ -7,117 +7,249 @@ tags:
 # 京都大学 情報学研究科 知能情報学専攻 2022年8月実施 情報学基礎 F1-1
 
 ## **Author**
-[Isidore](https://github.com/heacsing)
+[Isidore](https://github.com/heacsing), 祭音Myyura
 
 ## **Description**
-<figure style="text-align:center;">
-  <img src="https://s2.loli.net/2024/06/27/KiB4xw9ncTDaLs7.png" width="640"/>
-</figure>
+### 設問1
+以下の行列 $A$ に対して、$A = LU$ を満たす下三角行列 $L$ と上三角行列 $U$ を求めよ。ただし $L$ の対角成分はすべて 1 とする。
+
+$$
+A = 
+\begin{pmatrix}
+  -6 & -9 & -2 & 7 & -9 \\
+  42 & 59 & 7 & -53 & 56 \\
+  30 & 37 & -8 & -35 & 30 \\
+  -42 & -47 & 30 & 35 & -33 \\
+  12 & 18 & 20 & -64 & 43
+\end{pmatrix}
+$$
+
+### 設問2
+四元数の実 4 次正方行列表現における基底元は以下のように定義される。
+
+$$
+E = 
+\begin{pmatrix}
+  1 & 0 & 0 & 0 \\
+  0 & 1 & 0 & 0 \\
+  0 & 0 & 1 & 0 \\
+  0 & 0 & 0 & 1
+\end{pmatrix}, \ 
+I = 
+\begin{pmatrix}
+  0 & 1 & 0 & 0 \\
+  -1 & 0 & 0 & 0 \\
+  0 & 0 & 0 & -1 \\
+  0 & 0 & 1 & 0
+\end{pmatrix}, 
+$$
+
+$$
+J = 
+\begin{pmatrix}
+  0 & 0 & 1 & 0 \\
+  0 & 0 & 0 & 1 \\
+  -1 & 0 & 0 & 0 \\
+  0 & -1 & 0 & 0
+\end{pmatrix}, \ 
+K = 
+\begin{pmatrix}
+  0 & 0 & 0 & 1 \\
+  0 & 0 & -1 & 0 \\
+  0 & 1 & 0 & 0 \\
+  -1 & 0 & 0 & 0
+\end{pmatrix}
+$$
+
+以下の問いに答えよ。次の等式を用いてもよい：
+
+$$
+IJ = K, \  JK = I, \  KI = J, \  JI = -K, \  KJ = -I, \  IK = -J,
+$$
+
+$$
+I^2 = J^2 = K^2 = IJK = -E
+$$
+
+(1) $(a, b, c, d) \in \mathbb{R}^4$ とし、$Q = aE + bI + cJ + dK$, $\overline{Q} = aE - bI - cJ - dK$ として $Q\overline{Q}$ を求めよ。
+
+(2) $I^{-1}$ と $Q^{-1}$ を求めよ。ただし $(a, b, c, d) \neq 0$ とする。
+
+(3) 実 4 次正方行列の集合 $M$ は非可換環である。この部分集合 $H = \{Q \mid \forall(a, b, c, d)\}$ も非可換環であるための以下の必要条件を証明せよ：
+   - (a) $H$ は加法に対して閉じている。
+   - (b) 加法交換則が成り立つ。
+   - ($c$) 加法結合則が成り立つ。
+   - (d) 加法に対する零元が存在する。
+   - (e) 加法に対する逆元が存在する。
+   - (f) $H$ は乗法に対して閉じている。
+   - (g) 乗法結合則が成り立つ。
+   - (h) 乗法分配則が成り立つ。
+   - (i) 乗法は非可換である。
 
 ## **Kai**
 ### 設問1
 
 $$
 A=
-\begin{bmatrix}
+\begin{pmatrix}
     1 & 0 & 0 & 0 & 0 \\
     -7 & 1 & 0 & 0 & 0 \\
     -5 & 2 & 1 & 0 & 0 \\
     7 & -4 & -4 & 1 & 0 \\
     -2 & 0 & -4 & -9 & 1
-\end{bmatrix}
-\begin{bmatrix}
+\end{pmatrix}
+\begin{pmatrix}
     -6 & -9 & -2 & 7 & -9 \\ 
-    0 & -4 & -7 & -4 & 7 \\ 
+    0 & -4 & -7 & -4 & -7 \\ 
     0 & 0 & -4 & 8 & -1 \\
     0 & 0 & 0 & 2 & -2 \\
     0 & 0 & 0 & 0 & 3 
-\end{bmatrix}
+\end{pmatrix}
 $$
 
 ### 設問2
 #### (1)
 
 $$
-Q\bar{Q} = (a^2+b^2+c^2+d^2)E 
+\begin{aligned}
+Q\overline{Q} &= (aE+bI+cJ+dK)(aE-bI-cJ-dK)\\
+&= (a^{2}E^{2}-abEI-acEJ-adEK) + (abIE-b^{2}I^{2}-bcIJ-bdIK)\\
+&\quad+(acJE-bcJI-c^{2}J^{2}-cdJK)+(adKE-bdKI-cdJK-d^{2}K^{2})\\
+&= (a^{2}E-abI-acJ-adK) + (abI+b^{2}-bcK+bdJ)\\
+&\quad+(acJ+bcK+c^{2}-cdI)+(adK-bdJ+cdI-d^{2})\\
+&= (a^{2}+b^{2}+c^{2}+d^{2})E
+\end{aligned}
 $$
 
 #### (2)
 
 $$
 I^2 = -E \Rightarrow I^{-1} = 
-\begin{bmatrix}
+\begin{pmatrix}
     0 & -1 & 0 & 0 \\
     1 & 0 & 0 & 0 \\
     0 & 0 & 0 & 1 \\
     0 & 0 & -1 & 0 \\
-\end{bmatrix}
+\end{pmatrix}
+= -I
 $$
 
 $$
-Q\bar{Q} = (a^2+b^2+c^2+d^2)E \Rightarrow Q^{-1} = \frac{1}{a^2+b^2+c^2+d^2}
-\begin{bmatrix}
+Q\overline{Q} = (a^2+b^2+c^2+d^2)E \Rightarrow Q^{-1} = \frac{1}{a^2+b^2+c^2+d^2}
+\begin{pmatrix}
     a & -b & -c & d \\
     b & a & d & -c \\
     c & -d & a & b \\
     d & c & -b & a \\
-\end{bmatrix}
+\end{pmatrix}
 $$
 
 #### (3)
+The complete proving is to use $(a, b, c, d)$ to represent all the $Q$s below with their calculations, which is easy but tedious, hence some proof is omitted.
+
+For $i = 1,2,3$, let $Q_{i} = a_{i}E+b_{i}I+c_{i}J+d_{i}K$.
 
-The complete proving is to use $(a, b, c, d)$ to represent all the $Q$s below with their calculations, which is easy but tedious.
+(a): $\forall Q_1, Q_2 \in H, Q_1 + Q_2 \in H$
 
-- (a)  
-
 $$
-\forall Q_1, Q_2 \in H, Q_1 + Q_2 \in H
+\begin{aligned}
+Q_{1}+Q_{2} &= (a_{1}E+b_{1}I+c_{1}J+d_{1}K)+(a_{2}E+b_{2}I+c_{2}J+d_{2}K)\\
+&= (a_{1}+a_{2})E+(b_{1}+b_{2})I+(c_{1}+c_{2})J+(d_{1}+d_{2})K\in H
+\end{aligned}
 $$
 
-- (b)
-  
+(b): $\forall Q_1, Q_2 \in H, Q_1 + Q_2 = Q_2 + Q_1$
+
 $$
-\forall Q_1, Q_2 \in H, Q_1 + Q_2 = Q_2 + Q_1
+\begin{aligned}
+Q_{1}+Q_{2} &= (a_{1}E+b_{1}I+c_{1}J+d_{1}K)+(a_{2}E+b_{2}I+c_{2}J+d_{2}K)\\
+&= (a_{2}E+b_{2}I+c_{2}J+d_{2}K)+(a_{1}E+b_{1}I+c_{1}J+d_{1}K) = Q_{2}+Q_{1}
+\end{aligned}
 $$
 
-- (c\) 
-  
+($c$): $\forall Q_1, Q_2, Q_3 \in H, (Q_1 + Q_2) + Q_3 = Q_1 + (Q_2 + Q_3)$
+
 $$
-\forall Q_1, Q_2, Q_3 \in H, (Q_1 + Q_2) + Q_3 = Q_1 + (Q_2 + Q_3)
+\begin{aligned}
+(Q_{1}+Q_{2})+Q_{3} &= \left\{(a_{1}E+b_{1}I+c_{1}J+d_{1}K)+(a_{2}E+b_{2}I+c_{2}J+d_{2}K)\right\}+(a_{3}E+b_{3}I+c_{3}J+d_{3}K)\\
+&= (a_{1}E+b_{1}I+c_{1}J+d_{1}K)+\left\{(a_{2}E+b_{2}I+c_{2}J+d_{2}K)+(a_{3}E+b_{3}I+c_{3}J+d_{3}K)\right\}\\
+&= Q_{1}+(Q_{2}+Q_{3})
+\end{aligned}
 $$
 
-- (d) 
-
+(d): $\exists O \in H,  \forall Q \in H, O + Q = Q$.
+
+Let $O$ denote the zero-martix. When $(a, b, c, d) = 0$ we have $Q_1 = O$, hence $O \in H$. And we have
+
 $$
-\exists Q_0 \in H,  \forall Q \in H, Q_0 + Q = Q
+Q_{1}+O = O+Q_{1} = Q_{1}
 $$
 
-- (e) 
-
+proof finishes.
+
+(e): $\forall Q \in H, \exists Q' \in H,  Q + Q' = O$.
+
+Let $Q_{1}^{\prime}=-Q_{1}$. Then,
+
 $$
-\forall Q \in H, \exists \hat{Q} \in H,  Q + \hat{Q} = Q_0
+\begin{aligned}
+Q_{1}^{\prime} &= -(a_{1}E+b_{1}I+c_{1}J+d_{1}K) \\
+&= (-a_{1})E+(-b_{1})I+(-c_{1})J+(-d_{1})K\in H
+\end{aligned}
 $$
 
-- (f) 
-
+since $Q_{1}+Q_{1}^{\prime}=Q_{1}^{\prime}Q_{1}=O$, the proof finishes.
+
+(f): $\forall Q_1, Q_2 \in H, Q_1Q_2 \in H$
+
 $$
-\forall Q_1, Q_2 \in H, Q_1Q_2 \in H
+\begin{aligned}
+Q_{1}Q_{2} &= (a_{1}E+b_{1}I+c_{1}J+d_{1}K)+(a_{2}E+b_{2}I+c_{2}J+d_{2}K) \\
+&= (a_{1}a_{2}E^{2}+a_{1}b_{2}EI+a_{1}c_{2}EJ+a_{1}d_{2}EK) + (a_{2}b_{1}IE+b_{1}b_{2}I^{2}+b_{1}c_{2}IJ+b_{1}d_{2}IK)\\
+&\quad+(a_{2}c_{1}JE+b_{2}c_{1}JI+c_{1}c_{2}J^{2}+c_{1}d_{2}JK)+(a_{2}d_{1}KE+b_{2}d_{1}KI+c_{2}d_{1}KJ+d_{1}d_{2}K^{2})\\
+&= (a_{1}a_{2}E+a_{1}b_{2}I+a_{1}c_{2}J+a_{1}d_{2}K) + (a_{2}b_{1}I-b_{1}b_{2}E+b_{1}c_{2}K-b_{1}d_{2}J)\\
+&\quad+(a_{2}c_{1}J-b_{2}c_{1}K-c_{1}c_{2}E+c_{1}d_{2}I)+(a_{2}d_{1}K+b_{2}d_{1}J-c_{2}d_{1}I-d_{1}d_{2}E)\\
+&= (a_{1}a_{2}-b_{1}b_{2}-c_{1}c_{2}-d_{1}d_{2})E+(a_{1}b_{2}+a_{2}b_{1}+c_{1}d_{2}-c_{2}d_{1})I\\
+&\quad+(a_{1}c_{2}+a_{2}c_{1}-b_{1}d_{2}+b_{2}d_{1})J+(a_{1}d_{2}+a_{2}d_{1}+b_{1}c_{2}-b_{2}c_{1})K\\
+&\equiv a^{\prime}E+b^{\prime}I+c^{\prime}J+d^{\prime}K\equiv Q^{\prime} \in H
+\end{aligned}
 $$
 
-- (g)
-
+(g): $\forall Q_1, Q_2, Q_3 \in H, (Q_1Q_2)Q_3 = Q_1(Q_2Q_3)$
+
+Omitted
+
+(h): $\forall Q_1, Q_2, Q_3 \in H, (Q_1 + Q_2)Q_3 = Q_1Q_3 + Q_2Q_3$
+
+Omitted
+
+(i): $\exists Q_1, Q_2 \in H, Q_1Q_2 \neq Q_2Q_1$
+
+let
+
 $$
-\forall Q_1, Q_2, Q_3 \in H, (Q_1Q_2)Q_3 = Q_1(Q_2Q_3)
+\begin{aligned}
+(a_{1},b_{1},c_{1},d_{1}) &= (0,0,2,1) \\
+(a_{2},b_{2},c_{2},d_{2}) &= (0,0,-1,2) \\
+\end{aligned}
 $$
 
-- (h) 
-
 $$
-\forall Q_1, Q_2, Q_3 \in H, (Q_1 + Q_2)Q_3 = Q_1Q_3 + Q_2Q_3
+\begin{aligned}
+Q_{1}Q_{2} &=
+(a_{1}a_{2}-b_{1}b_{2}-c_{1}c_{2}-d_{1}d_{2})E+(a_{1}b_{2}+a_{2}b_{1}+c_{1}d_{2}-c_{2}d_{1})I\\
+&\quad+(a_{1}c_{2}+a_{2}c_{1}-b_{1}d_{2}+b_{2}d_{1})J+(a_{1}d_{2}+a_{2}d_{1}+b_{1}c_{2}-b_{2}c_{1})K \\
+&= 5I
+\end{aligned}
 $$
 
-- (i) 
-
 $$
-\exists Q_1, Q_2 \in H, Q_1Q_2 \neq Q_2Q_1
+\begin{aligned}
+Q_{2}Q_{1} &=
+(a_{2}a_{1}-b_{2}b_{1}-c_{2}c_{1}-d_{2}d_{1})E+(a_{2}b_{1}+a_{1}b_{2}+c_{2}d_{1}-c_{1}d_{2})I\\
+&\quad+(a_{2}c_{1}+a_{1}c_{2}-b_{2}d_{1}+b_{1}d_{2})J+(a_{2}d_{1}+a_{1}d_{2}+b_{2}c_{1}-b_{1}c_{2})K \\
+&= -5I
+\end{aligned}
 $$
+
+proof finishes.