練習問題 メモ 3#
3.1#
\(A\) を \(k \times l\) 行列、 \(B\) を \(k \times n\) 行列、 \(C\) を \(m \times n\) 行列とする。分割された \((k+m) \times(l+n)\)行列に対する次の計算をせよ。 \( \left(\begin{array}{cc} A & B \\ O_{m, l} & C \end{array}\right)+\left(\begin{array}{cc} A & -B \\ O_{m, l} & -2 C \end{array}\right) \)
\[\begin{split}
\left(\begin{array}{cc}
A & B \\
O_{m, l} & C
\end{array}\right)
+
\left(\begin{array}{cc}
A & -B \\
O_{m, l} & -2 C
\end{array}\right)
=
\left(\begin{array}{cc}
A+A & B-B \\
O_{m, l}-O_{m, l} & C-2 C
\end{array}\right)
=
\left(\begin{array}{cc}
2A & O_{k,n} \\
O_{m, l} & -C
\end{array}\right)
\end{split}\]
3.2#
\(A\) を \(m \times n\) 行列とする。分割された \((m+n)\) 次の正方行列 \(\left(\begin{array}{cc}E_m & A \\ O_{n, m} & E_n\end{array}\right)\)の3乗を計算せよ。
\[\begin{split}
\left(\begin{array}{cc}E_m & A \\ O_{n, m} & E_n\end{array}\right)^3
= \left(\begin{array}{cc}
E_m^3 & 3A \\
O_{n, m} & E_n^3
\end{array}\right)
= \left(\begin{array}{cc}
E_m & 3A \\
O_{n, m} & E_n
\end{array}\right)
\end{split}\]
\[\begin{split}
\underbrace{
\begin{pmatrix}
E_m & A \\
O_{n, m} & E_n
\end{pmatrix}
\begin{pmatrix}
E_m & A \\
O_{n, m} & E_n
\end{pmatrix}
}_{
\begin{pmatrix}
E_m & 2A \\
O_{n, m} & E_n
\end{pmatrix}
}
\begin{pmatrix}
E_m & A \\
O_{n, m} & E_n
\end{pmatrix}
=
\begin{pmatrix}
E_m & 3A \\
O_{n, m} & E_n
\end{pmatrix}
\end{split}\]
3.3#
4次の正方行列 \(I, J, K\)を $\( I=\left(\begin{array}{cccc} 0 & -1 & 0 & 0 \\ 1 & 0 & 0 & 0 \\ 0 & 0 & 0 & -1 \\ 0 & 0 & 1 & 0 \end{array}\right), J=\left(\begin{array}{cccc} 0 & 0 & -1 & 0 \\ 0 & 0 & 0 & 1 \\ 1 & 0 & 0 & 0 \\ 0 & -1 & 0 & 0 \end{array}\right), K=\left(\begin{array}{cccc} 0 & 0 & 0 & -1 \\ 0 & 0 & -1 & 0 \\ 0 & 1 & 0 & 0 \\ 1 & 0 & 0 & 0 \end{array}\right) \)\( により定める。 \)I, J, K\( を 2 次の正方行列を用いて分割することにより、積 \)I^2, J^2\(, \)K^2, I J, J I, J K, K J, K I, I K$ を計算せよ。
\[\begin{split}
A =
\begin{pmatrix}
0 & -1\\
1 & 0
\end{pmatrix}
,\hspace{1em}
B =
\begin{pmatrix}
-1 & 0\\
0 & 1
\end{pmatrix}
,\hspace{1em}
C =
\begin{pmatrix}
0 & 1\\
1 & 0
\end{pmatrix}
\end{split}\]
とおくと
\[\begin{split}
I=
\begin{pmatrix}
A & O\\
O & A
\end{pmatrix}, \
J=
\begin{pmatrix}
O & B\\
-B & O
\end{pmatrix}, \
K=
\begin{pmatrix}
O & -C\\
C & O
\end{pmatrix}
\end{split}\]
\(I^2\)
\[\begin{split}
I^2 =
\begin{pmatrix}
A & O\\
O & A
\end{pmatrix}
\begin{pmatrix}
A & O\\
O & A
\end{pmatrix}
=
\begin{pmatrix}
A^2 & O\\
O & A^2
\end{pmatrix}
\end{split}\]
\[\begin{split}
A^2 =
\begin{pmatrix}
0 & -1\\
1 & 0
\end{pmatrix}
\begin{pmatrix}
0 & -1\\
1 & 0
\end{pmatrix}
=
\begin{pmatrix}
-1 & 0\\
0 & -1
\end{pmatrix}
= - E_{2}
\end{split}\]
なので
\[\begin{split}
I^2 =
\begin{pmatrix}
-1 & 0 & 0 & 0 \\
0 & -1 & 0 & 0 \\
0 & 0 & -1 & 0 \\
0 & 0 & 0 & -1
\end{pmatrix}
\end{split}\]
Show code cell 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
Show code cell output
array([[-1, 0, 0, 0],
[ 0, -1, 0, 0],
[ 0, 0, -1, 0],
[ 0, 0, 0, -1]])
\(J^2\)
\[\begin{split}
J^2 =
\begin{pmatrix}
O & B\\
-B & O
\end{pmatrix}
\begin{pmatrix}
O & B\\
-B & O
\end{pmatrix}
=
\begin{pmatrix}
B (-B) & O\\
O & -B B
\end{pmatrix}
=
\begin{pmatrix}
-1 \cdot B^2 & O\\
O & -1 \cdot B^2
\end{pmatrix}
\end{split}\]
\[\begin{split}
B^2 =
\begin{pmatrix}
-1 & 0 \\
0 & 1
\end{pmatrix}
\begin{pmatrix}
-1 & 0 \\
0 & 1
\end{pmatrix}
=
\begin{pmatrix}
1 & 0 \\
0 & 1
\end{pmatrix}
= E_2
\end{split}\]
\[\begin{split}
J^2 =
\begin{pmatrix}
-1 & 0 & 0 & 0 \\
0 & -1 & 0 & 0 \\
0 & 0 & -1 & 0 \\
0 & 0 & 0 & -1
\end{pmatrix}
\end{split}\]
Show code cell 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
Show code cell output
array([[-1, 0, 0, 0],
[ 0, -1, 0, 0],
[ 0, 0, -1, 0],
[ 0, 0, 0, -1]])
\(K^2\)
\[\begin{split}
K^2 =
\begin{pmatrix}
O & -C\\
C & O
\end{pmatrix}
\begin{pmatrix}
O & -C\\
C & O
\end{pmatrix}
=
\begin{pmatrix}
-1\cdot C^2 & O\\
O & -1\cdot C^2
\end{pmatrix}
\end{split}\]
\[\begin{split}
C^2 =
\begin{pmatrix}
0 & 1 \\
1 & 0
\end{pmatrix}
\begin{pmatrix}
0 & 1 \\
1 & 0
\end{pmatrix}
=
\begin{pmatrix}
1 & 0 \\
0 & 1
\end{pmatrix}
= E_2
\end{split}\]
\[\begin{split}
K^2 =
\begin{pmatrix}
-1 & 0 & 0 & 0 \\
0 & -1 & 0 & 0 \\
0 & 0 & -1 & 0 \\
0 & 0 & 0 & -1
\end{pmatrix}
\end{split}\]
Show code cell 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
Show code cell output
array([[-1, 0, 0, 0],
[ 0, -1, 0, 0],
[ 0, 0, -1, 0],
[ 0, 0, 0, -1]])
\(IJ\)
\[\begin{split}
IJ
= \begin{pmatrix}
I_s & O\\
O & I_s
\end{pmatrix}
\begin{pmatrix}
O & J_1 \\
J_2 & O
\end{pmatrix}
= \begin{pmatrix}
O & I_s J_1 \\
I_s J_2 & O
\end{pmatrix}
\end{split}\]
\[\begin{split}
I_s J_1 =
\begin{pmatrix}
0 & -1 \\
1 & 0
\end{pmatrix}
\begin{pmatrix}
-1 & 0 \\
0 & 1
\end{pmatrix}
=
\begin{pmatrix}
0 & -1 \\
-1 & 0
\end{pmatrix}
\end{split}\]
\[\begin{split}
I_s J_2 =
\begin{pmatrix}
0 & -1 \\
1 & 0
\end{pmatrix}
\begin{pmatrix}
1 & 0 \\
0 & -1
\end{pmatrix}
=
\begin{pmatrix}
0 & 1 \\
1 & 0
\end{pmatrix}
\end{split}\]
\[\begin{split}
IJ = \begin{pmatrix}
0 & 0 & 0 & -1 \\
0 & 0 & -1 & 0 \\
0 & 1 & 0 & 0 \\
1 & 0 & 0 & 0
\end{pmatrix}
\end{split}\]
Show code cell source
I @ J
Show code cell output
array([[ 0, 0, 0, -1],
[ 0, 0, -1, 0],
[ 0, 1, 0, 0],
[ 1, 0, 0, 0]])
\(J I\)
\[\begin{split}
JI
=
\begin{pmatrix}
O & J_1 \\
J_2 & O
\end{pmatrix}
\begin{pmatrix}
I_s & O\\
O & I_s
\end{pmatrix}
= \begin{pmatrix}
O & J_1 I_s\\
J_2 I_s & O
\end{pmatrix}
\end{split}\]
\[\begin{split}
J_1 I_s =
\begin{pmatrix}
-1 & 0 \\
0 & 1
\end{pmatrix}
\begin{pmatrix}
0 & -1 \\
1 & 0
\end{pmatrix}
=
\begin{pmatrix}
0 & 1 \\
1 & 0
\end{pmatrix}
\end{split}\]
\[\begin{split}
J_2 I_s =
\begin{pmatrix}
1 & 0 \\
0 & -1
\end{pmatrix}
\begin{pmatrix}
0 & -1 \\
1 & 0
\end{pmatrix}
=
\begin{pmatrix}
0 & -1 \\
-1 & 0
\end{pmatrix}
\end{split}\]
\[\begin{split}
JI =
\begin{pmatrix}
0 & 0 & 0 & 1 \\
0 & 0 & 1 & 0 \\
0 & -1 & 0 & 0 \\
-1 & 0 & 0 & 0
\end{pmatrix}
\end{split}\]
Show code cell source
J @ I
Show code cell output
array([[ 0, 0, 0, 1],
[ 0, 0, 1, 0],
[ 0, -1, 0, 0],
[-1, 0, 0, 0]])
\(J K\)
\[\begin{split}
J K =
\begin{pmatrix}
O & J_1 \\
J_2 & O
\end{pmatrix}
\begin{pmatrix}
O & K_1\\
K_2 & O
\end{pmatrix}
=
\begin{pmatrix}
J_1 K_2 & O\\
O & J_2 K_1
\end{pmatrix}
\end{split}\]
\[\begin{split}
J_1 K_2 =
\begin{pmatrix}
-1 & 0\\
0 & 1
\end{pmatrix}
\begin{pmatrix}
0 & 1\\
1 & 0
\end{pmatrix}
=
\begin{pmatrix}
0 & -1\\
1 & 0
\end{pmatrix}
\end{split}\]
\[\begin{split}
J_2 K_1 =
\begin{pmatrix}
1 & 0\\
0 & -1
\end{pmatrix}
\begin{pmatrix}
0 & -1\\
-1 & 0
\end{pmatrix}
=
\begin{pmatrix}
0 & -1\\
1 & 0
\end{pmatrix}
\end{split}\]
\[\begin{split}
JK =
\begin{pmatrix}
0 & -1 & 0 & 0 \\
1 & 0 & 0 & 0 \\
0 & 0 & 0 & -1 \\
0 & 0 & 1 & 0
\end{pmatrix}
\end{split}\]
Show code cell source
J @ K
Show code cell output
array([[ 0, -1, 0, 0],
[ 1, 0, 0, 0],
[ 0, 0, 0, -1],
[ 0, 0, 1, 0]])
\(K J\)
\[\begin{split}
K J =
\begin{pmatrix}
O & K_1\\
K_2 & O
\end{pmatrix}
\begin{pmatrix}
O & J_1 \\
J_2 & O
\end{pmatrix}
=
\begin{pmatrix}
K_1 J_2 & O\\
O & K_2 J_1
\end{pmatrix}
\end{split}\]
\[\begin{split}
K_2 J_1 =
\begin{pmatrix}
0 & 1\\
1 & 0
\end{pmatrix}
\begin{pmatrix}
-1 & 0\\
0 & 1
\end{pmatrix}
=
\begin{pmatrix}
0 & 1\\
-1 & 0\\
\end{pmatrix}
\end{split}\]
\[\begin{split}
K_1 J_2 =
\begin{pmatrix}
0 & -1\\
-1 & 0
\end{pmatrix}
\begin{pmatrix}
1 & 0\\
0 & -1
\end{pmatrix}
=
\begin{pmatrix}
0 & 1\\
-1 & 0
\end{pmatrix}
\end{split}\]
\[\begin{split}
KJ =
\begin{pmatrix}
0 & 1 & 0 & 0 \\
-1 & 0 & 0 & 0 \\
0 & 0 & 0 & 1 \\
0 & 0 & -1 & 0
\end{pmatrix}
\end{split}\]
Show code cell source
K @ J
Show code cell output
array([[ 0, 1, 0, 0],
[-1, 0, 0, 0],
[ 0, 0, 0, 1],
[ 0, 0, -1, 0]])
\(K I\)
\[\begin{split}
K I =
\begin{pmatrix}
O & K_1\\
K_2 & O
\end{pmatrix}
\begin{pmatrix}
I_s & O \\
O & I_s
\end{pmatrix}
=
\begin{pmatrix}
O & K_1 I_s\\
K_2 I_s & O
\end{pmatrix}
\end{split}\]
\[\begin{split}
K_1 I_s =
\begin{pmatrix}
0 & -1\\
-1 & 0
\end{pmatrix}
\begin{pmatrix}
0 & -1\\
1 & 0
\end{pmatrix}
=
\begin{pmatrix}
-1 & 0\\
0 & 1
\end{pmatrix}
\end{split}\]
\[\begin{split}
K_2 I_s =
\begin{pmatrix}
0 & 1\\
1 & 0
\end{pmatrix}
\begin{pmatrix}
0 & -1\\
1 & 0
\end{pmatrix}
=
\begin{pmatrix}
1 & 0\\
0 & -1\\
\end{pmatrix}
\end{split}\]
\[\begin{split}
KI =
\begin{pmatrix}
0 & 0 & -1 & 0 \\
0 & 0 & 0 & 1 \\
1 & 0 & 0 & 0 \\
0 & -1 & 0 & 0
\end{pmatrix}
\end{split}\]
Show code cell source
K @ I
Show code cell output
array([[ 0, 0, -1, 0],
[ 0, 0, 0, 1],
[ 1, 0, 0, 0],
[ 0, -1, 0, 0]])
\(I K\)
\[\begin{split}
I K =
\begin{pmatrix}
I_s & O \\
O & I_s
\end{pmatrix}
\begin{pmatrix}
O & K_1\\
K_2 & O
\end{pmatrix}
=
\begin{pmatrix}
O & I_s K_1\\
I_s K_2 & O
\end{pmatrix}
\end{split}\]
\[\begin{split}
I_s K_1 =
\begin{pmatrix}
0 & -1\\
1 & 0
\end{pmatrix}
\begin{pmatrix}
0 & -1\\
-1 & 0
\end{pmatrix}
=
\begin{pmatrix}
1 & 0 \\
0 & -1
\end{pmatrix}
\end{split}\]
\[\begin{split}
I_s K_2 =
\begin{pmatrix}
0 & -1\\
1 & 0
\end{pmatrix}
\begin{pmatrix}
0 & 1\\
1 & 0
\end{pmatrix}
=
\begin{pmatrix}
-1 & 0\\
0 & 1
\end{pmatrix}
\end{split}\]
\[\begin{split}
IK =
\begin{pmatrix}
0 & 0 & 1 & 0 \\
0 & 0 & 0 & -1 \\
-1 & 0 & 0 & 0 \\
0 & 1 & 0 & 0
\end{pmatrix}
\end{split}\]
Show code cell source
I @ K
Show code cell output
array([[ 0, 0, 1, 0],
[ 0, 0, 0, -1],
[-1, 0, 0, 0],
[ 0, 1, 0, 0]])
3.4#
\(A_{12}, A_{13}, A_{22}, A_{23}, A_{33}, B_{11}, B_{12}, B_{13}, B_{23}, B_{33}, C_{11}, C_{12}, C_{13}, C_{22}, C_{23}\) を\(n\)次の正方行列、\(O\)を\(n\)次の零行列とする。次の計算をせよ。 \( \left(\begin{array}{ccc} O & A_{12} & A_{13} \\ O & A_{22} & A_{23} \\ O & O & A_{33} \end{array}\right)\left(\begin{array}{ccc} B_{11} & B_{12} & B_{13} \\ O & O & B_{23} \\ O & O & B_{33} \end{array}\right)\left(\begin{array}{ccc} C_{11} & C_{12} & C_{13} \\ O & C_{22} & C_{23} \\ O & O & O \end{array}\right) \)
\[\begin{split}
\begin{pmatrix}
O & A_{12} & A_{13} \\
O & A_{22} & A_{23} \\
O & O & A_{33}
\end{pmatrix}
\begin{pmatrix}
B_{11} & B_{12} & B_{13} \\
O & O & B_{23} \\
O & O & B_{33}
\end{pmatrix}
=
\begin{pmatrix}
O & O & A_{12} B_{23} + A_{13} B_{33}\\
O & O & A_{22} B_{23} + A_{23} B_{33}\\
O & O & A_{33} B_{33}
\end{pmatrix}
\end{split}\]
\[\begin{split}
\begin{pmatrix}
O & O & A_{12} B_{23} + A_{13} B_{33}\\
O & O & A_{22} B_{23} + A_{23} B_{33}\\
O & O & A_{33} B_{33}
\end{pmatrix}
\begin{pmatrix}
C_{11} & C_{12} & C_{13} \\
O & C_{22} & C_{23} \\
O & O & O
\end{pmatrix}
=
\begin{pmatrix}
O & O & O\\
O & O & O\\
O & O & O
\end{pmatrix}
= O
\end{split}\]
Show code cell 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]
])
Show code cell source
A @ B
Show code cell output
\[\begin{split}\displaystyle \left[\begin{matrix}0 & 0 & A_{12} B_{23} + A_{13} B_{33}\\0 & 0 & A_{22} B_{23} + A_{23} B_{33}\\0 & 0 & A_{33} B_{33}\end{matrix}\right]\end{split}\]
Show code cell source
A @ B @ C
Show code cell output
\[\begin{split}\displaystyle \left[\begin{matrix}0 & 0 & 0\\0 & 0 & 0\\0 & 0 & 0\end{matrix}\right]\end{split}\]
3.5#
\(a, b\) を異なる数、 \(X_{11}, X_{12}, X_{21}, X_{22}\) をそれぞれ \(m\) 次の正方行列、 \(m \times n\) 行列、 \(n \times m\) 行列、 \(n\) 次の正方行列とし、 \((m+n)\) 次の正方行列 \(A\) および \(X\) を
\[\begin{split} A=\left(\begin{array}{cc} a E_m & O_{m, n} \\ O_{n, m} & b E_n \end{array}\right), X=\left(\begin{array}{ll} X_{11} & X_{12} \\ X_{21} & X_{22} \end{array}\right) \end{split}\]により定める。 \(A\) と \(X\) が可換となるのは \(X_{12}\) および \(X_{21}\) が零行列のときであることを示せ。
\[\begin{split}
AX =
\begin{pmatrix}
a E_m & O_{m, n} \\
O_{n, m} & b E_n
\end{pmatrix}
\begin{pmatrix}
X_{11} & X_{12} \\
X_{21} & X_{22}
\end{pmatrix}
=
\begin{pmatrix}
a E_m X_{11} & a E_m X_{12} \\
b E_n X_{21} & b E_n X_{22}
\end{pmatrix}
=
\begin{pmatrix}
a X_{11} & a X_{12} \\
b X_{21} & b X_{22}
\end{pmatrix}
\end{split}\]
\[\begin{split}
XA=
\begin{pmatrix}
X_{11} & X_{12} \\
X_{21} & X_{22}
\end{pmatrix}
\begin{pmatrix}
a E_m & O_{m, n} \\
O_{n, m} & b E_n
\end{pmatrix}
=
\begin{pmatrix}
a X_{11} E_m & b X_{12} E_n\\
a X_{21} E_m & b X_{22} E_n
\end{pmatrix}
=
\begin{pmatrix}
a X_{11} & b X_{12}\\
a X_{21} & b X_{22}
\end{pmatrix}
\end{split}\]
となり、\(a, b\)が異なる数であるため、\(AX\)と\(XA\)の対角成分は異なる値になっている。
\(X_{12}\)および\(X_{21}\)が零行列であれば対角成分は零行列になるため、\(AX = XA\)となる