練習問題 メモ 4

4.1

次の１～３の行列の階数標準形および階数(rank)を求めよ

1.

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

行ベクトルを

\[\begin{split} \begin{pmatrix} a_1 \\ a_2 \\ a_3 \end{pmatrix} \end{split}\]

とする

1. \(a_2' = a_2 + (-2) \times a_1\)

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]])

2.

\[\begin{split} \left(\begin{array}{lll}0 & 1 & 0 \\ 1 & 0 & 1 \\ 0 & 1 & 0\end{array}\right) \end{split}\]

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]])

3.

\[\begin{split} \left(\begin{array}{lll}a & 1 & 1 \\ 1 & a & 1 \\ 1 & 1 & a\end{array}\right) \end{split}\]

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

\[\begin{split}\displaystyle \left[\begin{matrix}a & 1 & 1\\1 & a & 1\\1 & 1 & a\end{matrix}\right]\end{split}\]

# 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

\[\begin{split}\displaystyle \left[\begin{matrix}1 & a & 1\\a & 1 & 1\\1 & 1 & a\end{matrix}\right]\end{split}\]

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

\[\begin{split}\displaystyle \left[\begin{matrix}1 & a & 1\\0 & 1 - a^{2} & 1 - a\\1 & 1 & a\end{matrix}\right]\end{split}\]

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

\[\begin{split}\displaystyle \left[\begin{matrix}1 & a & 1\\0 & 1 - a^{2} & 1 - a\\0 & 1 - a & a - 1\end{matrix}\right]\end{split}\]

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

\[\begin{split} \begin{pmatrix} 1 & a & 1\\ 0 & 1 - a^{2} & 1 - a\\ 0 & 1 - a & a - 1 \end{pmatrix} = \begin{pmatrix} 1 & 1 & 1\\ 0 & 1 - 1^{2} & 1 - 1\\ 0 & 1 - 1 & 1 - 1 \end{pmatrix} = \begin{pmatrix} 1 & 1 & 1\\ 0 & 0 & 0\\ 0 & 0 & 0 \end{pmatrix} \end{split}\]

列基本変形（列で引き算）して

\[\begin{split} \begin{pmatrix} 1 & 0 & 0\\ 0 & 0 & 0\\ 0 & 0 & 0 \end{pmatrix} \end{split}\]

→ rank=1

もし\(a \neq 1\)なら

\[\begin{split} \begin{pmatrix} 1 & a & 1\\ 0 & 1 - a^{2} & 1 - a\\ 0 & 1 - a & a - 1 \end{pmatrix} \end{split}\]

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

\[\begin{split}\displaystyle \left[\begin{matrix}1 & a & a + 1\\0 & 1 - a^{2} & - a^{2} - a + 2\\0 & 1 - a & 0\end{matrix}\right]\end{split}\]

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

\[\begin{split}\displaystyle \left[\begin{matrix}1 & -1 & a + 1\\0 & a - 1 & - a^{2} - a + 2\\0 & 1 - a & 0\end{matrix}\right]\end{split}\]

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

\[\begin{split}\displaystyle \left[\begin{matrix}1 & -1 & a + 1\\0 & 0 & - a^{2} - a + 2\\0 & 1 - a & 0\end{matrix}\right]\end{split}\]

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

\[\begin{split}\displaystyle \left[\begin{matrix}1 & -1 & a + 1\\0 & 1 - a & 0\\0 & 0 & - a^{2} - a + 2\end{matrix}\right]\end{split}\]

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

\[\begin{split}\displaystyle \left[\begin{matrix}1 & -1 & a + 1\\0 & 1 - a & - a^{2} - a + 2\\0 & 0 & - a^{2} - a + 2\end{matrix}\right]\end{split}\]

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

\[\begin{split}\displaystyle \left[\begin{matrix}1 & -1 & a\\0 & 1 - a & - a^{2} - 2 a + 3\\0 & 0 & - a^{2} - a + 2\end{matrix}\right]\end{split}\]

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

\[\begin{split}\displaystyle \left[\begin{matrix}1 & - a - 1 & a - 1\\0 & 1 - a & - a^{2} - 2 a + 3\\0 & 0 & - a^{2} - a + 2\end{matrix}\right]\end{split}\]

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

\[\begin{split}\displaystyle \left[\begin{matrix}1 & - a - 1 & a - 1\\0 & 1 - a & 1 - a\\0 & 0 & - a^{2} - a + 2\end{matrix}\right]\end{split}\]

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

\[\begin{split}\displaystyle \left[\begin{matrix}1 & - a - 1 & a - 1\\0 & \frac{1 - a}{a} & \frac{1 - a}{a}\\0 & 0 & - a^{2} - a + 2\end{matrix}\right]\end{split}\]

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

\[\begin{split}\displaystyle \left[\begin{matrix}1 & - a - 1 & a - 1\\0 & 1 - a & -2 + \frac{2}{a}\\0 & 0 & - a^{2} - a + 2\end{matrix}\right]\end{split}\]

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

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

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

\[\begin{split}\displaystyle \left[\begin{matrix}1 & -2 & a - 1\\0 & - a - 1 + \frac{2}{a} & -2 + \frac{2}{a}\\0 & - a^{2} - a + 2 & - a^{2} - a + 2\end{matrix}\right]\end{split}\]

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

\[\begin{split}\displaystyle \left[\begin{matrix}1 & -2 & a - 1\\0 & - a - 1 + \frac{2}{a} & -2 + \frac{2}{a}\\0 & - a^{2} - a - \frac{\left(1 - a\right) \left(- a - 1 + \frac{2}{a}\right)}{1 - a^{2}} + 2 & - a^{2} - a - \frac{\left(-2 + \frac{2}{a}\right) \left(1 - a\right)}{1 - a^{2}} + 2\end{matrix}\right]\end{split}\]

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

\[\begin{split}\displaystyle \left[\begin{matrix}1 & -2 & a - 1\\0 & - a - 1 + \frac{2}{a} & -2 + \frac{2}{a}\\0 & \frac{\left(a + 1\right) \left(- a^{2} - a - \frac{\left(1 - a\right) \left(- a - 1 + \frac{2}{a}\right)}{1 - a^{2}} + 2\right)}{\left(a - 1\right) \left(a + 2\right)} & \frac{\left(a + 1\right) \left(- a^{2} - a - \frac{\left(-2 + \frac{2}{a}\right) \left(1 - a\right)}{1 - a^{2}} + 2\right)}{\left(a - 1\right) \left(a + 2\right)}\end{matrix}\right]\end{split}\]

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

\[\begin{split}\displaystyle \left[\begin{matrix}1 & 0 & a + 1\\0 & 1 & -1\\0 & 0 & a + 2\end{matrix}\right]\end{split}\]

\(a = -2\)のとき

\(a \neq -2\)のとき

掃き出し法計算機 - Wolfram|Alpha

4.2

次の問に答えよ。

1. 2次の正方行列に対する階数標準形をすべて書け

\[\begin{split} \begin{pmatrix} 0 & 0 \\ 0 & 0 \\ \end{pmatrix} = O_2 , \hspace{1em} \begin{pmatrix} 1 & 0 \\ 0 & 0 \\ \end{pmatrix} , \hspace{1em} \begin{pmatrix} 1 & 0 \\ 0 & 1 \\ \end{pmatrix} = E_2 \end{split}\]

2. 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

\[\begin{split} \begin{pmatrix} 0 & 0 \\ 0 & 0 \\ \end{pmatrix} , \hspace{1em} \begin{pmatrix} 1 & * \\ 0 & 0 \\ \end{pmatrix} , \hspace{1em} \begin{pmatrix} 1 & *\\ 0 & 1 \\ \end{pmatrix} \end{split}\]