練習問題メモ 4

練習問題メモ 4#

次の１～３の行列の階数標準形および階数(rank)を求めよ

$(\begin{array}{ll} 1 & 2 \\ 3 & 4 \\ 5 & 6 \end{array})$

行ベクトルを

\begin{array}{r} (\begin{array}{c} a_{1} \\ a_{2} \\ a_{3} \end{array}) \end{array}

とする

import numpy as np
A = np.array([
    [1, 2],
    [3, 4],
    [5, 6],
])

# 1行目を-2倍して2行目に加える
A[1, :] += A[0, :] * (-2)
A

array([[1, 2],
       [1, 0],
       [5, 6]])

# 1行目を-3倍して3行目に加える
A[2, :] += A[0, :] * (-3)
A

array([[1, 2],
       [1, 0],
       [2, 0]])

# 3行目を(1/2)倍する
A[2, :] = (1 / 2) * A[2, :]
A

array([[1, 2],
       [1, 0],
       [1, 0]])

# 2行目を-1倍して3行目に加える
A[2, :] += A[1, :] * (-1)
A

array([[1, 2],
       [1, 0],
       [0, 0]])

# 2行目を-1倍して1行目に加える
A[0, :] += A[1, :] * (-1)
A

array([[0, 2],
       [1, 0],
       [0, 0]])

# 1行目を(1/2)倍する
A[0, :] = (1 / 2) * A[0, :]
A

array([[0, 1],
       [1, 0],
       [0, 0]])

# 1行目と2行目を入れ替える
a1 = A[0, :].copy()
a2 = A[1, :].copy()
A[0, :] = a2
A[1, :] = a1
A

array([[1, 0],
       [0, 1],
       [0, 0]])

$\begin{array}{r} (\begin{array}{lll} 0 & 1 & 0 \\ 1 & 0 & 1 \\ 0 & 1 & 0 \end{array}) \end{array}$

import numpy as np
A = np.array([
    [0, 1, 0],
    [1, 0, 1],
    [0, 1, 0],
])

# 1行目を-1倍して3行目に加える
A[2, :] += (-1) * A[0, :]
A

array([[0, 1, 0],
       [1, 0, 1],
       [0, 0, 0]])

# 0列目を-1倍して3列目に加える
A[:, 2] += (-1) * A[:, 0]
A

array([[0, 1, 0],
       [1, 0, 0],
       [0, 0, 0]])

# 1行目と2行目を入れ替える
a1 = A[0, :].copy()
a2 = A[1, :].copy()
A[0, :] = a2
A[1, :] = a1
A

array([[1, 0, 0],
       [0, 1, 0],
       [0, 0, 0]])

$\begin{array}{r} (\begin{array}{lll} a & 1 & 1 \\ 1 & a & 1 \\ 1 & 1 & a \end{array}) \end{array}$

わからなかった

from sympy import Symbol, Matrix

a = Symbol("a")
A = Matrix([
    [a, 1, 1],
    [1, a, 1],
    [1, 1, a],
])
A

\begin{array}{r} [\begin{array}{c} a & 1 & 1 \\ 1 & a & 1 \\ 1 & 1 & a \end{array}] \end{array}

# 1行目と2行目を入れ替える
a1 = A[0, :].copy()
a2 = A[1, :].copy()
A[0, :] = a2
A[1, :] = a1
A

\begin{array}{r} [\begin{array}{c} 1 & a & 1 \\ a & 1 & 1 \\ 1 & 1 & a \end{array}] \end{array}

A[1, :] += -a * A[0, :]
A

\begin{array}{r} [\begin{array}{c} 1 & a & 1 \\ 0 & 1 - a^{2} & 1 - a \\ 1 & 1 & a \end{array}] \end{array}

A[2, :] += -1 * A[0, :]
A

\begin{array}{r} [\begin{array}{c} 1 & a & 1 \\ 0 & 1 - a^{2} & 1 - a \\ 0 & 1 - a & a - 1 \end{array}] \end{array}

未知の変数があって最終的なランクが変わりうる → 場合分けする

もし $a = 1$ なら

\begin{array}{r} (\begin{array}{c} 1 & a & 1 \\ 0 & 1 - a^{2} & 1 - a \\ 0 & 1 - a & a - 1 \end{array}) = (\begin{array}{c} 1 & 1 & 1 \\ 0 & 1 - 1^{2} & 1 - 1 \\ 0 & 1 - 1 & 1 - 1 \end{array}) = (\begin{array}{c} 1 & 1 & 1 \\ 0 & 0 & 0 \\ 0 & 0 & 0 \end{array}) \end{array}

列基本変形（列で引き算）して

\begin{array}{r} (\begin{array}{c} 1 & 0 & 0 \\ 0 & 0 & 0 \\ 0 & 0 & 0 \end{array}) \end{array}

→ rank=1

もし $a \neq 1$ なら

\begin{array}{r} (\begin{array}{c} 1 & a & 1 \\ 0 & 1 - a^{2} & 1 - a \\ 0 & 1 - a & a - 1 \end{array}) \end{array}

A[:, 2] += 1 * A[:, 1]
A

\begin{array}{r} [\begin{array}{c} 1 & a & a + 1 \\ 0 & 1 - a^{2} & - a^{2} - a + 2 \\ 0 & 1 - a & 0 \end{array}] \end{array}

A[:, 1] += -1 * A[:, 2]
A

\begin{array}{r} [\begin{array}{c} 1 & - 1 & a + 1 \\ 0 & a - 1 & - a^{2} - a + 2 \\ 0 & 1 - a & 0 \end{array}] \end{array}

A[1, :] += A[2, :]
A

\begin{array}{r} [\begin{array}{c} 1 & - 1 & a + 1 \\ 0 & 0 & - a^{2} - a + 2 \\ 0 & 1 - a & 0 \end{array}] \end{array}

a1 = A[1, :].copy()
a2 = A[2, :].copy()
A[1, :] = a2
A[2, :] = a1
A

\begin{array}{r} [\begin{array}{c} 1 & - 1 & a + 1 \\ 0 & 1 - a & 0 \\ 0 & 0 & - a^{2} - a + 2 \end{array}] \end{array}

A[1, :] += A[2, :]
A

\begin{array}{r} [\begin{array}{c} 1 & - 1 & a + 1 \\ 0 & 1 - a & - a^{2} - a + 2 \\ 0 & 0 & - a^{2} - a + 2 \end{array}] \end{array}

A[:, 2] += 1 * A[:, 1]
A

\begin{array}{r} [\begin{array}{c} 1 & - 1 & a \\ 0 & 1 - a & - a^{2} - 2 a + 3 \\ 0 & 0 & - a^{2} - a + 2 \end{array}] \end{array}

A[:, 1] += -a * A[:, 0]
A[:, 2] += -1 * A[:, 0]
A

\begin{array}{r} [\begin{array}{c} 1 & - a - 1 & a - 1 \\ 0 & 1 - a & - a^{2} - 2 a + 3 \\ 0 & 0 & - a^{2} - a + 2 \end{array}] \end{array}

A[1, :] += -1 * A[2, :]
A

\begin{array}{r} [\begin{array}{c} 1 & - a - 1 & a - 1 \\ 0 & 1 - a & 1 - a \\ 0 & 0 & - a^{2} - a + 2 \end{array}] \end{array}

A[1, :] = (1/a) * A[1, :]
A

\begin{array}{r} [\begin{array}{c} 1 & - a - 1 & a - 1 \\ 0 & \frac{1 - a}{a} & \frac{1 - a}{a} \\ 0 & 0 & - a^{2} - a + 2 \end{array}] \end{array}

A[1, 1] = -a + 1
A[1, 2] = (2/a) - 2
A

\begin{array}{r} [\begin{array}{c} 1 & - a - 1 & a - 1 \\ 0 & 1 - a & - 2 + \frac{2}{a} \\ 0 & 0 & - a^{2} - a + 2 \end{array}] \end{array}

A[2, :] - A[1, :]

[\begin{matrix} 0 & a - 1 & - a^{2} - a + 4 - \frac{2}{a} \end{matrix}]

A[:, 1] += 1 * A[:, 2]
A

\begin{array}{r} [\begin{array}{c} 1 & - 2 & a - 1 \\ 0 & - a - 1 + \frac{2}{a} & - 2 + \frac{2}{a} \\ 0 & - a^{2} - a + 2 & - a^{2} - a + 2 \end{array}] \end{array}

A[2, :] += -((1-a)/(1-a**2)) * A[1, :]
A

\begin{array}{r} [\begin{array}{c} 1 & - 2 & a - 1 \\ 0 & - a - 1 + \frac{2}{a} & - 2 + \frac{2}{a} \\ 0 & - a^{2} - a - \frac{(1 - a) (- a - 1 + \frac{2}{a})}{1 - a^{2}} + 2 & - a^{2} - a - \frac{(- 2 + \frac{2}{a}) (1 - a)}{1 - a^{2}} + 2 \end{array}] \end{array}

A[2, :] = (1+a)/((-1+a)*(2+a)) * A[2, :]
A

\begin{array}{r} [\begin{array}{c} 1 & - 2 & a - 1 \\ 0 & - a - 1 + \frac{2}{a} & - 2 + \frac{2}{a} \\ 0 & \frac{(a + 1) (- a^{2} - a - \frac{(1 - a) (- a - 1 + \frac{2}{a})}{1 - a^{2}} + 2)}{(a - 1) (a + 2)} & \frac{(a + 1) (- a^{2} - a - \frac{(- 2 + \frac{2}{a}) (1 - a)}{1 - a^{2}} + 2)}{(a - 1) (a + 2)} \end{array}] \end{array}

Matrix([
[1, 0, 1+a],
[0 ,1, -1],
[0, 0, 2+a]
])

\begin{array}{r} [\begin{array}{c} 1 & 0 & a + 1 \\ 0 & 1 & - 1 \\ 0 & 0 & a + 2 \end{array}] \end{array}

$a = - 2$ のとき

$a \neq - 2$ のとき

次の問に答えよ。

2次の正方行列に対する階数標準形をすべて書け

\begin{array}{r} (\begin{array}{c} 0 & 0 \\ 0 & 0 \end{array}) = O_{2}, (\begin{array}{c} 1 & 0 \\ 0 & 0 \end{array}), (\begin{array}{c} 1 & 0 \\ 0 & 1 \end{array}) = E_{2} \end{array}

2次の正方行列に対する階段行列をすべて書け。なお、任意の数は*を用いて表せ。

$* = 0$ の状況も考えて1とおくと

\begin{array}{r} (\begin{array}{c} 0 & 0 \\ 0 & 0 \end{array}), (\begin{array}{c} 1 & * \\ 0 & 0 \end{array}), (\begin{array}{c} 1 & * \\ 0 & 1 \end{array}) \end{array}