Kyoto-University
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 , 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.