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練習問題 メモ 3

3.1

AAk×lk \times l 行列、 BBk×nk \times n 行列、 CCm×nm \times n 行列とする。分割された (k+m)×(l+n)(k+m) \times(l+n)行列に対する次の計算をせよ。 (ABOm,lC)+(ABOm,l2C) \left(\begin{array}{cc} A & B \\ O_{m, l} & C \end{array}\right)+\left(\begin{array}{cc} A & -B \\ O_{m, l} & -2 C \end{array}\right)

(ABOm,lC)+(ABOm,l2C)=(A+ABBOm,lOm,lC2C)=(2AOk,nOm,lC)\left(\begin{array}{cc} A & B \\ O_{m, l} & C \end{array}\right) + \left(\begin{array}{cc} A & -B \\ O_{m, l} & -2 C \end{array}\right) = \left(\begin{array}{cc} A+A & B-B \\ O_{m, l}-O_{m, l} & C-2 C \end{array}\right) = \left(\begin{array}{cc} 2A & O_{k,n} \\ O_{m, l} & -C \end{array}\right)

3.2

AAm×nm \times n 行列とする。分割された (m+n)(m+n) 次の正方行列 (EmAOn,mEn)\left(\begin{array}{cc}E_m & A \\ O_{n, m} & E_n\end{array}\right)の3乗を計算せよ。

(EmAOn,mEn)3=(Em33AOn,mEn3)=(Em3AOn,mEn)\left(\begin{array}{cc}E_m & A \\ O_{n, m} & E_n\end{array}\right)^3 = \left(\begin{array}{cc} E_m^3 & 3A \\ O_{n, m} & E_n^3 \end{array}\right) = \left(\begin{array}{cc} E_m & 3A \\ O_{n, m} & E_n \end{array}\right)
(EmAOn,mEn)(EmAOn,mEn)(Em2AOn,mEn)(EmAOn,mEn)=(Em3AOn,mEn)\underbrace{ \begin{pmatrix} E_m & A \\ O_{n, m} & E_n \end{pmatrix} \begin{pmatrix} E_m & A \\ O_{n, m} & E_n \end{pmatrix} }_{ \begin{pmatrix} E_m & 2A \\ O_{n, m} & E_n \end{pmatrix} } \begin{pmatrix} E_m & A \\ O_{n, m} & E_n \end{pmatrix} = \begin{pmatrix} E_m & 3A \\ O_{n, m} & E_n \end{pmatrix}

3.3

4次の正方行列 I,J,KI, J, K

I=(0100100000010010),J=(0010000110000100),K=(0001001001001000)I=\left(\begin{array}{cccc} 0 & -1 & 0 & 0 \\ 1 & 0 & 0 & 0 \\ 0 & 0 & 0 & -1 \\ 0 & 0 & 1 & 0 \end{array}\right), J=\left(\begin{array}{cccc} 0 & 0 & -1 & 0 \\ 0 & 0 & 0 & 1 \\ 1 & 0 & 0 & 0 \\ 0 & -1 & 0 & 0 \end{array}\right), K=\left(\begin{array}{cccc} 0 & 0 & 0 & -1 \\ 0 & 0 & -1 & 0 \\ 0 & 1 & 0 & 0 \\ 1 & 0 & 0 & 0 \end{array}\right)

により定める。 I,J,KI, J, K を 2 次の正方行列を用いて分割することにより、積 I2,J2I^2, J^2, K2,IJ,JI,JK,KJ,KI,IKK^2, I J, J I, J K, K J, K I, I K を計算せよ。

A=(0110),B=(1001),C=(0110)A = \begin{pmatrix} 0 & -1\\ 1 & 0 \end{pmatrix} ,\hspace{1em} B = \begin{pmatrix} -1 & 0\\ 0 & 1 \end{pmatrix} ,\hspace{1em} C = \begin{pmatrix} 0 & 1\\ 1 & 0 \end{pmatrix}

とおくと

I=(AOOA), J=(OBBO), K=(OCCO)I= \begin{pmatrix} A & O\\ O & A \end{pmatrix}, \ J= \begin{pmatrix} O & B\\ -B & O \end{pmatrix}, \ K= \begin{pmatrix} O & -C\\ C & O \end{pmatrix}

I2I^2

I2=(AOOA)(AOOA)=(A2OOA2)I^2 = \begin{pmatrix} A & O\\ O & A \end{pmatrix} \begin{pmatrix} A & O\\ O & A \end{pmatrix} = \begin{pmatrix} A^2 & O\\ O & A^2 \end{pmatrix}
A2=(0110)(0110)=(1001)=E2A^2 = \begin{pmatrix} 0 & -1\\ 1 & 0 \end{pmatrix} \begin{pmatrix} 0 & -1\\ 1 & 0 \end{pmatrix} = \begin{pmatrix} -1 & 0\\ 0 & -1 \end{pmatrix} = - E_{2}

なので

I2=(1000010000100001)I^2 = \begin{pmatrix} -1 & 0 & 0 & 0 \\ 0 & -1 & 0 & 0 \\ 0 & 0 & -1 & 0 \\ 0 & 0 & 0 & -1 \end{pmatrix}
Source
import numpy as np
I = np.array([
    [0, -1, 0, 0],
    [1, 0, 0, 0],
    [0, 0, 0, -1],
    [0, 0, 1, 0]
])

I @ I
Output
array([[-1, 0, 0, 0], [ 0, -1, 0, 0], [ 0, 0, -1, 0], [ 0, 0, 0, -1]])

J2J^2

J2=(OBBO)(OBBO)=(B(B)OOBB)=(1B2OO1B2)J^2 = \begin{pmatrix} O & B\\ -B & O \end{pmatrix} \begin{pmatrix} O & B\\ -B & O \end{pmatrix} = \begin{pmatrix} B (-B) & O\\ O & -B B \end{pmatrix} = \begin{pmatrix} -1 \cdot B^2 & O\\ O & -1 \cdot B^2 \end{pmatrix}
B2=(1001)(1001)=(1001)=E2B^2 = \begin{pmatrix} -1 & 0 \\ 0 & 1 \end{pmatrix} \begin{pmatrix} -1 & 0 \\ 0 & 1 \end{pmatrix} = \begin{pmatrix} 1 & 0 \\ 0 & 1 \end{pmatrix} = E_2
J2=(1000010000100001)J^2 = \begin{pmatrix} -1 & 0 & 0 & 0 \\ 0 & -1 & 0 & 0 \\ 0 & 0 & -1 & 0 \\ 0 & 0 & 0 & -1 \end{pmatrix}
Source
import numpy as np
J = np.array([
    [0, 0, -1, 0],
    [0, 0, 0, 1],
    [1, 0, 0, 0],
    [0, -1, 0, 0]
])

J @ J
Output
array([[-1, 0, 0, 0], [ 0, -1, 0, 0], [ 0, 0, -1, 0], [ 0, 0, 0, -1]])

K2K^2

K2=(OCCO)(OCCO)=(1C2OO1C2)K^2 = \begin{pmatrix} O & -C\\ C & O \end{pmatrix} \begin{pmatrix} O & -C\\ C & O \end{pmatrix} = \begin{pmatrix} -1\cdot C^2 & O\\ O & -1\cdot C^2 \end{pmatrix}
C2=(0110)(0110)=(1001)=E2C^2 = \begin{pmatrix} 0 & 1 \\ 1 & 0 \end{pmatrix} \begin{pmatrix} 0 & 1 \\ 1 & 0 \end{pmatrix} = \begin{pmatrix} 1 & 0 \\ 0 & 1 \end{pmatrix} = E_2
K2=(1000010000100001)K^2 = \begin{pmatrix} -1 & 0 & 0 & 0 \\ 0 & -1 & 0 & 0 \\ 0 & 0 & -1 & 0 \\ 0 & 0 & 0 & -1 \end{pmatrix}
Source
import numpy as np
K = np.array([
    [0, 0, 0, -1],
    [0, 0, -1, 0],
    [0, 1, 0, 0],
    [1, 0, 0, 0]
])

K @ K
Output
array([[-1, 0, 0, 0], [ 0, -1, 0, 0], [ 0, 0, -1, 0], [ 0, 0, 0, -1]])

IJIJ

IJ=(IsOOIs)(OJ1J2O)=(OIsJ1IsJ2O)IJ = \begin{pmatrix} I_s & O\\ O & I_s \end{pmatrix} \begin{pmatrix} O & J_1 \\ J_2 & O \end{pmatrix} = \begin{pmatrix} O & I_s J_1 \\ I_s J_2 & O \end{pmatrix}
IsJ1=(0110)(1001)=(0110)I_s J_1 = \begin{pmatrix} 0 & -1 \\ 1 & 0 \end{pmatrix} \begin{pmatrix} -1 & 0 \\ 0 & 1 \end{pmatrix} = \begin{pmatrix} 0 & -1 \\ -1 & 0 \end{pmatrix}
IsJ2=(0110)(1001)=(0110)I_s J_2 = \begin{pmatrix} 0 & -1 \\ 1 & 0 \end{pmatrix} \begin{pmatrix} 1 & 0 \\ 0 & -1 \end{pmatrix} = \begin{pmatrix} 0 & 1 \\ 1 & 0 \end{pmatrix}
IJ=(0001001001001000)IJ = \begin{pmatrix} 0 & 0 & 0 & -1 \\ 0 & 0 & -1 & 0 \\ 0 & 1 & 0 & 0 \\ 1 & 0 & 0 & 0 \end{pmatrix}
Source
I @ J
Output
array([[ 0, 0, 0, -1], [ 0, 0, -1, 0], [ 0, 1, 0, 0], [ 1, 0, 0, 0]])

JIJ I

JI=(OJ1J2O)(IsOOIs)=(OJ1IsJ2IsO)JI = \begin{pmatrix} O & J_1 \\ J_2 & O \end{pmatrix} \begin{pmatrix} I_s & O\\ O & I_s \end{pmatrix} = \begin{pmatrix} O & J_1 I_s\\ J_2 I_s & O \end{pmatrix}
J1Is=(1001)(0110)=(0110)J_1 I_s = \begin{pmatrix} -1 & 0 \\ 0 & 1 \end{pmatrix} \begin{pmatrix} 0 & -1 \\ 1 & 0 \end{pmatrix} = \begin{pmatrix} 0 & 1 \\ 1 & 0 \end{pmatrix}
J2Is=(1001)(0110)=(0110)J_2 I_s = \begin{pmatrix} 1 & 0 \\ 0 & -1 \end{pmatrix} \begin{pmatrix} 0 & -1 \\ 1 & 0 \end{pmatrix} = \begin{pmatrix} 0 & -1 \\ -1 & 0 \end{pmatrix}
JI=(0001001001001000)JI = \begin{pmatrix} 0 & 0 & 0 & 1 \\ 0 & 0 & 1 & 0 \\ 0 & -1 & 0 & 0 \\ -1 & 0 & 0 & 0 \end{pmatrix}
Source
J @ I
Output
array([[ 0, 0, 0, 1], [ 0, 0, 1, 0], [ 0, -1, 0, 0], [-1, 0, 0, 0]])

JKJ K

JK=(OJ1J2O)(OK1K2O)=(J1K2OOJ2K1)J K = \begin{pmatrix} O & J_1 \\ J_2 & O \end{pmatrix} \begin{pmatrix} O & K_1\\ K_2 & O \end{pmatrix} = \begin{pmatrix} J_1 K_2 & O\\ O & J_2 K_1 \end{pmatrix}
J1K2=(1001)(0110)=(0110)J_1 K_2 = \begin{pmatrix} -1 & 0\\ 0 & 1 \end{pmatrix} \begin{pmatrix} 0 & 1\\ 1 & 0 \end{pmatrix} = \begin{pmatrix} 0 & -1\\ 1 & 0 \end{pmatrix}
J2K1=(1001)(0110)=(0110)J_2 K_1 = \begin{pmatrix} 1 & 0\\ 0 & -1 \end{pmatrix} \begin{pmatrix} 0 & -1\\ -1 & 0 \end{pmatrix} = \begin{pmatrix} 0 & -1\\ 1 & 0 \end{pmatrix}
JK=(0100100000010010)JK = \begin{pmatrix} 0 & -1 & 0 & 0 \\ 1 & 0 & 0 & 0 \\ 0 & 0 & 0 & -1 \\ 0 & 0 & 1 & 0 \end{pmatrix}
Source
J @ K
Output
array([[ 0, -1, 0, 0], [ 1, 0, 0, 0], [ 0, 0, 0, -1], [ 0, 0, 1, 0]])

KJK J

KJ=(OK1K2O)(OJ1J2O)=(K1J2OOK2J1)K J = \begin{pmatrix} O & K_1\\ K_2 & O \end{pmatrix} \begin{pmatrix} O & J_1 \\ J_2 & O \end{pmatrix} = \begin{pmatrix} K_1 J_2 & O\\ O & K_2 J_1 \end{pmatrix}
K2J1=(0110)(1001)=(0110)K_2 J_1 = \begin{pmatrix} 0 & 1\\ 1 & 0 \end{pmatrix} \begin{pmatrix} -1 & 0\\ 0 & 1 \end{pmatrix} = \begin{pmatrix} 0 & 1\\ -1 & 0\\ \end{pmatrix}
K1J2=(0110)(1001)=(0110)K_1 J_2 = \begin{pmatrix} 0 & -1\\ -1 & 0 \end{pmatrix} \begin{pmatrix} 1 & 0\\ 0 & -1 \end{pmatrix} = \begin{pmatrix} 0 & 1\\ -1 & 0 \end{pmatrix}
KJ=(0100100000010010)KJ = \begin{pmatrix} 0 & 1 & 0 & 0 \\ -1 & 0 & 0 & 0 \\ 0 & 0 & 0 & 1 \\ 0 & 0 & -1 & 0 \end{pmatrix}
Source
K @ J
Output
array([[ 0, 1, 0, 0], [-1, 0, 0, 0], [ 0, 0, 0, 1], [ 0, 0, -1, 0]])

KIK I

KI=(OK1K2O)(IsOOIs)=(OK1IsK2IsO)K I = \begin{pmatrix} O & K_1\\ K_2 & O \end{pmatrix} \begin{pmatrix} I_s & O \\ O & I_s \end{pmatrix} = \begin{pmatrix} O & K_1 I_s\\ K_2 I_s & O \end{pmatrix}
K1Is=(0110)(0110)=(1001)K_1 I_s = \begin{pmatrix} 0 & -1\\ -1 & 0 \end{pmatrix} \begin{pmatrix} 0 & -1\\ 1 & 0 \end{pmatrix} = \begin{pmatrix} -1 & 0\\ 0 & 1 \end{pmatrix}
K2Is=(0110)(0110)=(1001)K_2 I_s = \begin{pmatrix} 0 & 1\\ 1 & 0 \end{pmatrix} \begin{pmatrix} 0 & -1\\ 1 & 0 \end{pmatrix} = \begin{pmatrix} 1 & 0\\ 0 & -1\\ \end{pmatrix}
KI=(0010000110000100)KI = \begin{pmatrix} 0 & 0 & -1 & 0 \\ 0 & 0 & 0 & 1 \\ 1 & 0 & 0 & 0 \\ 0 & -1 & 0 & 0 \end{pmatrix}
Source
K @ I
Output
array([[ 0, 0, -1, 0], [ 0, 0, 0, 1], [ 1, 0, 0, 0], [ 0, -1, 0, 0]])

IKI K

IK=(IsOOIs)(OK1K2O)=(OIsK1IsK2O)I K = \begin{pmatrix} I_s & O \\ O & I_s \end{pmatrix} \begin{pmatrix} O & K_1\\ K_2 & O \end{pmatrix} = \begin{pmatrix} O & I_s K_1\\ I_s K_2 & O \end{pmatrix}
IsK1=(0110)(0110)=(1001)I_s K_1 = \begin{pmatrix} 0 & -1\\ 1 & 0 \end{pmatrix} \begin{pmatrix} 0 & -1\\ -1 & 0 \end{pmatrix} = \begin{pmatrix} 1 & 0 \\ 0 & -1 \end{pmatrix}
IsK2=(0110)(0110)=(1001)I_s K_2 = \begin{pmatrix} 0 & -1\\ 1 & 0 \end{pmatrix} \begin{pmatrix} 0 & 1\\ 1 & 0 \end{pmatrix} = \begin{pmatrix} -1 & 0\\ 0 & 1 \end{pmatrix}
IK=(0010000110000100)IK = \begin{pmatrix} 0 & 0 & 1 & 0 \\ 0 & 0 & 0 & -1 \\ -1 & 0 & 0 & 0 \\ 0 & 1 & 0 & 0 \end{pmatrix}
Source
I @ K
Output
array([[ 0, 0, 1, 0], [ 0, 0, 0, -1], [-1, 0, 0, 0], [ 0, 1, 0, 0]])

3.4

A12,A13,A22,A23,A33,B11,B12,B13,B23,B33,C11,C12,C13,C22,C23A_{12}, A_{13}, A_{22}, A_{23}, A_{33}, B_{11}, B_{12}, B_{13}, B_{23}, B_{33}, C_{11}, C_{12}, C_{13}, C_{22}, C_{23}nn次の正方行列、OOnn次の零行列とする。次の計算をせよ。 (OA12A13OA22A23OOA33)(B11B12B13OOB23OOB33)(C11C12C13OC22C23OOO) \left(\begin{array}{ccc} O & A_{12} & A_{13} \\ O & A_{22} & A_{23} \\ O & O & A_{33} \end{array}\right)\left(\begin{array}{ccc} B_{11} & B_{12} & B_{13} \\ O & O & B_{23} \\ O & O & B_{33} \end{array}\right)\left(\begin{array}{ccc} C_{11} & C_{12} & C_{13} \\ O & C_{22} & C_{23} \\ O & O & O \end{array}\right)

(OA12A13OA22A23OOA33)(B11B12B13OOB23OOB33)=(OOA12B23+A13B33OOA22B23+A23B33OOA33B33)\begin{pmatrix} O & A_{12} & A_{13} \\ O & A_{22} & A_{23} \\ O & O & A_{33} \end{pmatrix} \begin{pmatrix} B_{11} & B_{12} & B_{13} \\ O & O & B_{23} \\ O & O & B_{33} \end{pmatrix} = \begin{pmatrix} O & O & A_{12} B_{23} + A_{13} B_{33}\\ O & O & A_{22} B_{23} + A_{23} B_{33}\\ O & O & A_{33} B_{33} \end{pmatrix}
(OOA12B23+A13B33OOA22B23+A23B33OOA33B33)(C11C12C13OC22C23OOO)=(OOOOOOOOO)=O\begin{pmatrix} O & O & A_{12} B_{23} + A_{13} B_{33}\\ O & O & A_{22} B_{23} + A_{23} B_{33}\\ O & O & A_{33} B_{33} \end{pmatrix} \begin{pmatrix} C_{11} & C_{12} & C_{13} \\ O & C_{22} & C_{23} \\ O & O & O \end{pmatrix} = \begin{pmatrix} O & O & O\\ O & O & O\\ O & O & O \end{pmatrix} = O
Source
# 検算
from sympy import symbols, Matrix
A_12, A_13, A_22, A_23, A_33 = symbols("A_12, A_13, A_22, A_23, A_33")
B_11, B_12, B_13, B_23, B_33 = symbols("B_11, B_12, B_13, B_23, B_33")
C_11, C_12, C_13, C_22, C_23 = symbols("C_11, C_12, C_13, C_22, C_23")

A = Matrix([
    [0, A_12, A_13],
    [0, A_22, A_23],
    [0, 0, A_33]
])


B = Matrix([
    [B_11, B_12, B_13],
    [0, 0, B_23],
    [0, 0, B_33]
])

C = Matrix([
    [C_11, C_12, C_13],
    [0, C_22, C_23],
    [0, 0, 0]
])
Output
Source
A @ B
Output
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Source
A @ B @ C
Output
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3.5

a,ba, b を異なる数、 X11,X12,X21,X22X_{11}, X_{12}, X_{21}, X_{22} をそれぞれ mm 次の正方行列、 m×nm \times n 行列、 n×mn \times m 行列、 nn 次の正方行列とし、 (m+n)(m+n) 次の正方行列 AA および XX

A=(aEmOm,nOn,mbEn),X=(X11X12X21X22)A=\left(\begin{array}{cc} a E_m & O_{m, n} \\ O_{n, m} & b E_n \end{array}\right), X=\left(\begin{array}{ll} X_{11} & X_{12} \\ X_{21} & X_{22} \end{array}\right)

により定める。 AAXX が可換となるのは X12X_{12} および X21X_{21} が零行列のときであることを示せ。

AX=(aEmOm,nOn,mbEn)(X11X12X21X22)=(aEmX11aEmX12bEnX21bEnX22)=(aX11aX12bX21bX22)AX = \begin{pmatrix} a E_m & O_{m, n} \\ O_{n, m} & b E_n \end{pmatrix} \begin{pmatrix} X_{11} & X_{12} \\ X_{21} & X_{22} \end{pmatrix} = \begin{pmatrix} a E_m X_{11} & a E_m X_{12} \\ b E_n X_{21} & b E_n X_{22} \end{pmatrix} = \begin{pmatrix} a X_{11} & a X_{12} \\ b X_{21} & b X_{22} \end{pmatrix}
XA=(X11X12X21X22)(aEmOm,nOn,mbEn)=(aX11EmbX12EnaX21EmbX22En)=(aX11bX12aX21bX22)XA= \begin{pmatrix} X_{11} & X_{12} \\ X_{21} & X_{22} \end{pmatrix} \begin{pmatrix} a E_m & O_{m, n} \\ O_{n, m} & b E_n \end{pmatrix} = \begin{pmatrix} a X_{11} E_m & b X_{12} E_n\\ a X_{21} E_m & b X_{22} E_n \end{pmatrix} = \begin{pmatrix} a X_{11} & b X_{12}\\ a X_{21} & b X_{22} \end{pmatrix}

となり、a,ba, bが異なる数であるため、AXAXXAXAの対角成分は異なる値になっている。

X12X_{12}およびX21X_{21}が零行列であれば対角成分は零行列になるため、AX=XAAX = XAとなる