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

标签：

• Kyoto-University
• Lower-Upper-Decomposition
• Quaternion-Algebra

Author

Isidore, 祭音Myyura

Description

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

$$A= \begin{pmatrix} 1 & 0 & 0 & 0 & 0 \\ -7 & 1 & 0 & 0 & 0 \\ -5 & 2 & 1 & 0 & 0 \\ 7 & -4 & -4 & 1 & 0 \\ -2 & 0 & -4 & -9 & 1 \end{pmatrix} \begin{pmatrix} -6 & -9 & -2 & 7 & -9 \\ 0 & -4 & -7 & -4 & -7 \\ 0 & 0 & -4 & 8 & -1 \\ 0 & 0 & 0 & 2 & -2 \\ 0 & 0 & 0 & 0 & 3 \end{pmatrix}$$

設問2

(1)

$$\begin{aligned} Q\overline{Q} &= (aE+bI+cJ+dK)(aE-bI-cJ-dK)\\ &= (a^{2}E^{2}-abEI-acEJ-adEK) + (abIE-b^{2}I^{2}-bcIJ-bdIK)\\ &\quad+(acJE-bcJI-c^{2}J^{2}-cdJK)+(adKE-bdKI-cdJK-d^{2}K^{2})\\ &= (a^{2}E-abI-acJ-adK) + (abI+b^{2}-bcK+bdJ)\\ &\quad+(acJ+bcK+c^{2}-cdI)+(adK-bdJ+cdI-d^{2})\\ &= (a^{2}+b^{2}+c^{2}+d^{2})E \end{aligned}$$

(2)

$$I^2 = -E \Rightarrow I^{-1} = \begin{pmatrix} 0 & -1 & 0 & 0 \\ 1 & 0 & 0 & 0 \\ 0 & 0 & 0 & 1 \\ 0 & 0 & -1 & 0 \\ \end{pmatrix} = -I$$

$$Q\overline{Q} = (a^2+b^2+c^2+d^2)E \Rightarrow Q^{-1} = \frac{1}{a^2+b^2+c^2+d^2} \begin{pmatrix} a & -b & -c & d \\ b & a & d & -c \\ c & -d & a & b \\ d & c & -b & a \\ \end{pmatrix}$$

(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 $Q_{i} = a_{i}E+b_{i}I+c_{i}J+d_{i}K$.

(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): $\forall Q_1, Q_2 \in H, Q_1 + Q_2 = Q_2 + Q_1$

$$\begin{aligned} Q_{1}+Q_{2} &= (a_{1}E+b_{1}I+c_{1}J+d_{1}K)+(a_{2}E+b_{2}I+c_{2}J+d_{2}K)\\ &= (a_{2}E+b_{2}I+c_{2}J+d_{2}K)+(a_{1}E+b_{1}I+c_{1}J+d_{1}K) = Q_{2}+Q_{1} \end{aligned}$$

($c$): $\forall Q_1, Q_2, Q_3 \in H, (Q_1 + Q_2) + Q_3 = Q_1 + (Q_2 + Q_3)$

$$\begin{aligned} (Q_{1}+Q_{2})+Q_{3} &= \left\{(a_{1}E+b_{1}I+c_{1}J+d_{1}K)+(a_{2}E+b_{2}I+c_{2}J+d_{2}K)\right\}+(a_{3}E+b_{3}I+c_{3}J+d_{3}K)\\ &= (a_{1}E+b_{1}I+c_{1}J+d_{1}K)+\left\{(a_{2}E+b_{2}I+c_{2}J+d_{2}K)+(a_{3}E+b_{3}I+c_{3}J+d_{3}K)\right\}\\ &= Q_{1}+(Q_{2}+Q_{3}) \end{aligned}$$

(d): $\exists O \in H, \forall Q \in H, O + Q = Q$.

Let $O$ denote the zero-martix. When $(a, b, c, d) = 0$ we have $Q_1 = O$, hence $O \in H$. And we have

$$Q_{1}+O = O+Q_{1} = Q_{1}$$

proof finishes.

(e): $\forall Q \in H, \exists Q' \in H, Q + Q' = O$.

Let $Q_{1}^{\prime}=-Q_{1}$. Then,

$$\begin{aligned} Q_{1}^{\prime} &= -(a_{1}E+b_{1}I+c_{1}J+d_{1}K) \\ &= (-a_{1})E+(-b_{1})I+(-c_{1})J+(-d_{1})K\in H \end{aligned}$$

since $Q_{1}+Q_{1}^{\prime}=Q_{1}^{\prime}Q_{1}=O$, the proof finishes.

(f): $\forall Q_1, Q_2 \in H, Q_1Q_2 \in H$

$$\begin{aligned} Q_{1}Q_{2} &= (a_{1}E+b_{1}I+c_{1}J+d_{1}K)+(a_{2}E+b_{2}I+c_{2}J+d_{2}K) \\ &= (a_{1}a_{2}E^{2}+a_{1}b_{2}EI+a_{1}c_{2}EJ+a_{1}d_{2}EK) + (a_{2}b_{1}IE+b_{1}b_{2}I^{2}+b_{1}c_{2}IJ+b_{1}d_{2}IK)\\ &\quad+(a_{2}c_{1}JE+b_{2}c_{1}JI+c_{1}c_{2}J^{2}+c_{1}d_{2}JK)+(a_{2}d_{1}KE+b_{2}d_{1}KI+c_{2}d_{1}KJ+d_{1}d_{2}K^{2})\\ &= (a_{1}a_{2}E+a_{1}b_{2}I+a_{1}c_{2}J+a_{1}d_{2}K) + (a_{2}b_{1}I-b_{1}b_{2}E+b_{1}c_{2}K-b_{1}d_{2}J)\\ &\quad+(a_{2}c_{1}J-b_{2}c_{1}K-c_{1}c_{2}E+c_{1}d_{2}I)+(a_{2}d_{1}K+b_{2}d_{1}J-c_{2}d_{1}I-d_{1}d_{2}E)\\ &= (a_{1}a_{2}-b_{1}b_{2}-c_{1}c_{2}-d_{1}d_{2})E+(a_{1}b_{2}+a_{2}b_{1}+c_{1}d_{2}-c_{2}d_{1})I\\ &\quad+(a_{1}c_{2}+a_{2}c_{1}-b_{1}d_{2}+b_{2}d_{1})J+(a_{1}d_{2}+a_{2}d_{1}+b_{1}c_{2}-b_{2}c_{1})K\\ &\equiv a^{\prime}E+b^{\prime}I+c^{\prime}J+d^{\prime}K\equiv Q^{\prime} \in H \end{aligned}$$

(g): $\forall Q_1, Q_2, Q_3 \in H, (Q_1Q_2)Q_3 = Q_1(Q_2Q_3)$

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

$$\begin{aligned} (a_{1},b_{1},c_{1},d_{1}) &= (0,0,2,1) \\ (a_{2},b_{2},c_{2},d_{2}) &= (0,0,-1,2) \\ \end{aligned}$$

$$\begin{aligned} Q_{1}Q_{2} &= (a_{1}a_{2}-b_{1}b_{2}-c_{1}c_{2}-d_{1}d_{2})E+(a_{1}b_{2}+a_{2}b_{1}+c_{1}d_{2}-c_{2}d_{1})I\\ &\quad+(a_{1}c_{2}+a_{2}c_{1}-b_{1}d_{2}+b_{2}d_{1})J+(a_{1}d_{2}+a_{2}d_{1}+b_{1}c_{2}-b_{2}c_{1})K \\ &= 5I \end{aligned}$$

$$\begin{aligned} Q_{2}Q_{1} &= (a_{2}a_{1}-b_{2}b_{1}-c_{2}c_{1}-d_{2}d_{1})E+(a_{2}b_{1}+a_{1}b_{2}+c_{2}d_{1}-c_{1}d_{2})I\\ &\quad+(a_{2}c_{1}+a_{1}c_{2}-b_{2}d_{1}+b_{1}d_{2})J+(a_{2}d_{1}+a_{1}d_{2}+b_{2}c_{1}-b_{1}c_{2})K \\ &= -5I \end{aligned}$$

proof finishes.