練習問題メモ 24（対称行列の対角化）

練習問題メモ 24（対称行列の対角化）#

問1#

次の対称行列 $A$ を直交行列によって対角化せよ。

$A = (\begin{array}{ll} 3 & 2 \\ 2 & 6 \end{array})$
$A = (\begin{array}{lll} 1 & 2 & 3 \\ 2 & 4 & 6 \\ 3 & 6 & 9 \end{array})$ ただし、 $A$ の固有値 $λ$ が $λ = 0$ (重解) 14 で、それぞれの

固有値に対する固有空間が

\begin{array}{r} \begin{aligned} W (0) & = {c_{1} (\begin{array}{c} - 2 \\ 1 \\ 0 \end{array}) + c_{2} (\begin{array}{c} - 3 \\ 0 \\ 1 \end{array}) | c_{1}, c_{2} \in R} \\ W (14) & = {c (\begin{matrix} 1 \\ 2 \\ 3 \end{matrix}) | c \in R} \end{aligned} \end{array}

であることを用いてよい。

$A = (\begin{array}{ll} 3 & 2 \\ 2 & 6 \end{array})$

まず固有値を求める

\begin{array}{r} \begin{aligned} | A - λ E | & = (3 - λ) (6 - λ) - 2 \cdot 2 \\ = 18 - 9 λ + λ^{2} - 4 \\ = 14 - 9 λ + λ^{2} \\ = (λ - 2) (λ - 7) \end{aligned} \end{array}

よって $λ = 2, 7$

続いて、固有ベクトルを求める。

$λ = 2$ の場合、

\begin{array}{r} (A - λ E) x = 0 \\ ⟺ (\begin{array}{c} 3 - λ & 2 \\ 2 & 6 - λ \end{array}) (\begin{array}{c} x_{1} \\ x_{2} \end{array}) = (\begin{array}{c} 0 \\ 0 \end{array}) \\ ⟺ {\begin{cases} x_{1} + 2 x_{2} = 0 \\ 2 x_{1} + 4 x_{2} = 0 \end{cases} \end{array}

よって $x_{1} = - 2 x_{2}$ より、 $c \in R$ を用いて

\begin{array}{r} (\begin{array}{c} x_{1} \\ x_{2} \end{array}) = c (\begin{array}{c} - 2 \\ 1 \end{array}) \end{array}

と表すことができる。固有空間は

\begin{array}{r} V (2) = {c (\begin{array}{c} - 2 \\ 1 \end{array}) | c \in R} \end{array}

となる。

$λ = 7$ の場合、

\begin{array}{r} (\begin{array}{c} 3 - λ & 2 \\ 2 & 6 - λ \end{array}) (\begin{array}{c} x_{1} \\ x_{2} \end{array}) = (\begin{array}{c} 0 \\ 0 \end{array}) \\ ⟺ {\begin{cases} - 4 x_{1} + 2 x_{2} = 0 \\ 2 x_{1} - x_{2} = 0 \end{cases} \end{array}

よって $2 x_{1} = x_{2}$ より、 $c \in R$ を用いて

\begin{array}{r} (\begin{array}{c} x_{1} \\ x_{2} \end{array}) = c (\begin{array}{c} 1 \\ 2 \end{array}) \end{array}

と表すことができる。固有空間は

\begin{array}{r} V (7) = {c (\begin{array}{l} 1 \\ 2 \end{array}) | c \in R} \end{array}

となる。

固有値 $λ = 2, 7$ それぞれに属する固有ベクトル $x_{1}, x_{2}$ を

\begin{array}{r} x_{1} = (\begin{array}{c} - 2 \\ 1 \end{array}), x_{2} = (\begin{array}{c} 1 \\ 2 \end{array}) \end{array}

とおく。 $x_{1}, x_{2}$ にシュミットの直交化を行い $p_{1}, p_{2}$ とすると、

\begin{array}{r} \begin{aligned} p_{1} & = \frac{x_{1}}{‖ x_{1} ‖} = \frac{1}{\sqrt{5}} (\begin{array}{c} - 2 \\ 1 \end{array}) = (\begin{array}{c} \frac{- 2}{\sqrt{5}} \\ \frac{1}{\sqrt{5}} \end{array}) \end{aligned} \end{array}

であり

c = ⟨ x_{2}, p_{1} ⟩ = 1 \cdot \frac{- 2}{\sqrt{5}} + 2 \cdot \frac{1}{\sqrt{5}} = 0

より

\begin{array}{r} p_{2} = \frac{x_{2} - c p_{1}}{‖ x_{2} - c p_{1} ‖} = \frac{x_{2}}{‖ x_{2} ‖} = \frac{1}{\sqrt{5}} (\begin{array}{c} 1 \\ 2 \end{array}) = (\begin{array}{c} \frac{1}{\sqrt{5}} \\ \frac{2}{\sqrt{5}} \end{array}) \end{array}

$P = (p_{1}, p_{2})$ とおいた行列

\begin{array}{r} P = (\begin{array}{c} \frac{- 2}{\sqrt{5}} & \frac{1}{\sqrt{5}} \\ \frac{1}{\sqrt{5}} & \frac{2}{\sqrt{5}} \end{array}) \end{array}

により $A$ は対角化できる

\begin{array}{r} \begin{aligned} P^{- 1} A P & = (\begin{array}{c} \frac{- 2}{\sqrt{5}} & \frac{1}{\sqrt{5}} \\ \frac{1}{\sqrt{5}} & \frac{2}{\sqrt{5}} \end{array}) (\begin{array}{c} 3 & 2 \\ 2 & 6 \end{array}) (\begin{array}{c} \frac{- 2}{\sqrt{5}} & \frac{1}{\sqrt{5}} \\ \frac{1}{\sqrt{5}} & \frac{2}{\sqrt{5}} \end{array}) \\ = (\begin{array}{c} \frac{- 4}{\sqrt{5}} & \frac{2}{\sqrt{5}} \\ \frac{7}{\sqrt{5}} & \frac{14}{\sqrt{5}} \end{array}) (\begin{array}{c} \frac{- 2}{\sqrt{5}} & \frac{1}{\sqrt{5}} \\ \frac{1}{\sqrt{5}} & \frac{2}{\sqrt{5}} \end{array}) \\ = (\begin{array}{c} \frac{10}{5} & \frac{- 4 + 4}{5} \\ \frac{- 14 + 14}{5} & \frac{35}{5} \end{array}) \\ = (\begin{array}{c} 2 & 0 \\ 0 & 7 \end{array}) \end{aligned} \end{array}

import sympy as sp

x1 = sp.Matrix([-2,1])
x2 = sp.Matrix([1,2])
A = sp.Matrix([
    [3, 2],
    [2, 6],
])

p1 = x1 / x1.norm()
c = x2.T @ p1
v2 = (x2 - c[0]*p1)
p2 = v2 / v2.norm()
P = sp.Matrix([p1, p2]).reshape(2, 2).T
P

\begin{array}{r} [\begin{array}{c} - \frac{2 \sqrt{5}}{5} & \frac{\sqrt{5}}{5} \\ \frac{\sqrt{5}}{5} & \frac{2 \sqrt{5}}{5} \end{array}] \end{array}

P.inv() @ A @ P

\begin{array}{r} [\begin{array}{c} 2 & 0 \\ 0 & 7 \end{array}] \end{array}

P = sp.Matrix(sp.GramSchmidt([x1, x2])).reshape(2,2).T
P

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

P.inv() @ A @ P

\begin{array}{r} [\begin{array}{c} 2 & 0 \\ 0 & 7 \end{array}] \end{array}

$A = (\begin{array}{lll} 1 & 2 & 3 \\ 2 & 4 & 6 \\ 3 & 6 & 9 \end{array})$ ただし、 $A$ の固有値 $λ$ が $λ = 0$ (重解) 14 で、それぞれの

固有値に対する固有空間が

\begin{array}{r} \begin{aligned} W (0) & = {c_{1} (\begin{array}{c} - 2 \\ 1 \\ 0 \end{array}) + c_{2} (\begin{array}{c} - 3 \\ 0 \\ 1 \end{array}) | c_{1}, c_{2} \in R} \\ W (14) & = {c (\begin{matrix} 1 \\ 2 \\ 3 \end{matrix}) | c \in R} \end{aligned} \end{array}

であることを用いてよい。

固有ベクトルとして

\begin{array}{r} x_{1} = (\begin{array}{c} - 2 \\ 1 \\ 0 \end{array}), x_{2} = (\begin{array}{c} - 3 \\ 0 \\ 1 \end{array}), x_{3} = (\begin{array}{c} 1 \\ 2 \\ 3 \end{array}) \end{array}

を用いる。

これらを直交化したベクトルを $p_{1}, p_{2}, p_{3}$ とする。シュミットの直交化で求めていく。

(1) $p_{1}$ を求める

$‖ x_{1} ‖ = \sqrt{| 2 |^{2} + | 1 |^{2}} = \sqrt{5}$ なので、

\begin{array}{r} p_{1} = \frac{x_{1}}{‖ x_{1} ‖} = \frac{1}{\sqrt{5}} (\begin{array}{c} - 2 \\ 1 \\ 0 \end{array}) = (\begin{array}{c} \frac{- 2}{\sqrt{5}} \\ \frac{1}{\sqrt{5}} \\ 0 \end{array}) \end{array}

import sympy as sp

x1 = sp.Matrix([-2,1,0])
x2 = sp.Matrix([-3,0,1])
x3 = sp.Matrix([1,2,3])

p1 = x1 / x1.norm()
p1

\begin{array}{r} [\begin{array}{c} - \frac{2 \sqrt{5}}{5} \\ \frac{\sqrt{5}}{5} \\ 0 \end{array}] \end{array}

(2) $p_{2}$ を求める

c = ⟨ x_{2}, p_{1} ⟩ = - 3 \cdot \frac{- 2}{\sqrt{5}} = \frac{6}{\sqrt{5}}

\begin{array}{r} x_{2} - c p_{1} = (\begin{array}{c} - 3 \\ 0 \\ 1 \end{array}) - \frac{6}{\sqrt{5}} (\begin{array}{c} \frac{- 2}{\sqrt{5}} \\ \frac{1}{\sqrt{5}} \\ 0 \end{array}) = (\begin{array}{c} - 3 \\ 0 \\ 1 \end{array}) - (\begin{array}{c} \frac{- 12}{5} \\ \frac{6}{5} \\ 0 \end{array}) = (\begin{array}{c} \frac{- 15}{5} - \frac{- 12}{5} \\ - \frac{6}{5} \\ 1 \end{array}) = (\begin{array}{c} - \frac{3}{5} \\ - \frac{6}{5} \\ 1 \end{array}) \end{array}

‖ x_{2} - c p_{1} ‖ = \sqrt{{| - \frac{3}{5} |}^{2} + {| - \frac{6}{5} |}^{2} + 1^{2}} = \sqrt{\frac{9}{25} + \frac{36}{25} + \frac{25}{25}} = \sqrt{\frac{70}{25}} = \frac{\sqrt{70}}{5}

\begin{array}{r} p_{2} = \frac{x_{2} - c p_{1}}{‖ x_{2} - c p_{1} ‖} = \frac{5}{\sqrt{70}} (\begin{array}{c} - \frac{3}{5} \\ - \frac{6}{5} \\ 1 \end{array}) = (\begin{array}{c} - \frac{3}{\sqrt{70}} \\ - \frac{6}{\sqrt{70}} \\ \frac{5}{\sqrt{70}} \end{array}) \end{array}

c = (x2.T @ p1)[0]
p2 = (x2 - c * p1) / (x2 - c * p1).norm()
p2

\begin{array}{r} [\begin{array}{c} - \frac{3 \sqrt{70}}{70} \\ - \frac{3 \sqrt{70}}{35} \\ \frac{\sqrt{70}}{14} \end{array}] \end{array}

(3) $p_{3}$ を求める

\begin{array}{r} \begin{aligned} c_{1} & = ⟨ x_{3}, p_{1} ⟩ = 1 \cdot \frac{- 2}{\sqrt{5}} + 2 \cdot \frac{1}{\sqrt{5}} + 3 \cdot 0 = 0 \\ c_{2} & = ⟨ x_{3}, p_{2} ⟩ = 1 \cdot \frac{- 3}{\sqrt{70}} + 2 \cdot \frac{- 6}{\sqrt{70}} + 3 \cdot \frac{5}{\sqrt{70}} = \frac{- 3 - 12 + 15}{\sqrt{70}} = 0 \\ ‖ x_{3} ‖ & = \sqrt{| 1 |^{2} + | 2 |^{2} + | 3 |^{2}} = \sqrt{14} \end{aligned} \end{array}

なので

\begin{array}{r} p_{3} = \frac{x_{3} - (c_{1} p_{1} + c_{2} p_{2})}{‖ x_{3} - (c_{1} p_{1} + c_{2} p_{2}) ‖} = \frac{x_{3}}{‖ x_{3} ‖} = \frac{1}{\sqrt{14}} (\begin{array}{c} 1 \\ 2 \\ 3 \end{array}) \end{array}

$P = (p_{1}, p_{2}, p_{3})$ とすると、この $P$

\begin{array}{r} P = (\begin{array}{c} - \frac{2}{\sqrt{5}} & - \frac{3}{\sqrt{70}} & \frac{1}{\sqrt{14}} \\ \frac{1}{\sqrt{5}} & - \frac{6}{\sqrt{70}} & \frac{2}{\sqrt{14}} \\ 0 & \frac{5}{\sqrt{70}} & \frac{3}{\sqrt{14}} \end{array}) \end{array}

により $A$ は対角化できる

\begin{array}{r} P^{- 1} A P = [\begin{array}{ccc} 0 & 0 & 0 \\ 0 & 0 & 0 \\ 0 & 0 & 14 \end{array}] \end{array}

p3 = x3 / x3.norm()
p3

\begin{array}{r} [\begin{array}{c} \frac{\sqrt{14}}{14} \\ \frac{\sqrt{14}}{7} \\ \frac{3 \sqrt{14}}{14} \end{array}] \end{array}

P = sp.Matrix([p1, p2, p3]).reshape(3, 3).T
P

\begin{array}{r} [\begin{array}{c} - \frac{2 \sqrt{5}}{5} & - \frac{3 \sqrt{70}}{70} & \frac{\sqrt{14}}{14} \\ \frac{\sqrt{5}}{5} & - \frac{3 \sqrt{70}}{35} & \frac{\sqrt{14}}{7} \\ 0 & \frac{\sqrt{70}}{14} & \frac{3 \sqrt{14}}{14} \end{array}] \end{array}

P.inv()

\begin{array}{r} [\begin{array}{c} - \frac{2 \sqrt{5}}{5} & \frac{\sqrt{5}}{5} & 0 \\ - \frac{3 \sqrt{70}}{70} & - \frac{3 \sqrt{70}}{35} & \frac{\sqrt{70}}{14} \\ \frac{\sqrt{14}}{14} & \frac{\sqrt{14}}{7} & \frac{3 \sqrt{14}}{14} \end{array}] \end{array}

A = sp.Matrix([
    [1,2,3],
    [2,4,6],
    [3,6,9]
])

P.inv() @ A @ P

\begin{array}{r} [\begin{array}{c} 0 & 0 & 0 \\ 0 & 0 & 0 \\ 0 & 0 & 14 \end{array}] \end{array}

P = sp.Matrix(sp.GramSchmidt([x1,x2,x3], orthonormal=True)).reshape(3,3).T
P

\begin{array}{r} [\begin{array}{c} - \frac{2 \sqrt{5}}{5} & - \frac{3 \sqrt{70}}{70} & \frac{\sqrt{14}}{14} \\ \frac{\sqrt{5}}{5} & - \frac{3 \sqrt{70}}{35} & \frac{\sqrt{14}}{7} \\ 0 & \frac{\sqrt{70}}{14} & \frac{3 \sqrt{14}}{14} \end{array}] \end{array}

import sympy as sp

a1 = sp.Matrix([-2,1,0])
a2 = sp.Matrix([-3,0,1])
a3 = sp.Matrix([1,2,3])

p1 = a1 / a1.norm()
p2 = a2 / a2.norm()
p3 = a3 / a3.norm()
P = sp.Matrix([p1, p2, p3]).reshape(3, 3).T
P

\begin{array}{r} [\begin{array}{c} - \frac{2 \sqrt{5}}{5} & - \frac{3 \sqrt{10}}{10} & \frac{\sqrt{14}}{14} \\ \frac{\sqrt{5}}{5} & 0 & \frac{\sqrt{14}}{7} \\ 0 & \frac{\sqrt{10}}{10} & \frac{3 \sqrt{14}}{14} \end{array}] \end{array}

P.inv()

\begin{array}{r} [\begin{array}{c} - \frac{\sqrt{5}}{7} & \frac{5 \sqrt{5}}{7} & - \frac{3 \sqrt{5}}{7} \\ - \frac{3 \sqrt{10}}{14} & - \frac{3 \sqrt{10}}{7} & \frac{5 \sqrt{10}}{14} \\ \frac{\sqrt{14}}{14} & \frac{\sqrt{14}}{7} & \frac{3 \sqrt{14}}{14} \end{array}] \end{array}

A = sp.Matrix([
    [1,2,3],
    [2,4,6],
    [3,6,9]
])

P.inv() @ A @ P

\begin{array}{r} [\begin{array}{c} 0 & 0 & 0 \\ 0 & 0 & 0 \\ 0 & 0 & 14 \end{array}] \end{array}

Matrices (linear algebra) - SymPy 1.13.3 documentation

Gram–Schmidt process - Wikipedia

import sympy as sp

a1 = sp.Matrix([-2,1,0])
a2 = sp.Matrix([-3,0,1])
a3 = sp.Matrix([1,2,3])

P = sp.Matrix(sp.GramSchmidt([a1,a2,a3])).reshape(3,3).T
P

\begin{array}{r} [\begin{array}{c} - 2 & - \frac{3}{5} & 1 \\ 1 & - \frac{6}{5} & 2 \\ 0 & 1 & 3 \end{array}] \end{array}

P.inv() @ A @ P

\begin{array}{r} [\begin{array}{c} 0 & 0 & 0 \\ 0 & 0 & 0 \\ 0 & 0 & 14 \end{array}] \end{array}

補足：対称行列の対角化の応用例

実数を係数とする $x, y$ の 2 次方程式

a x^{2} + 2 b x y + c y^{2} + 2 d x + 2 e y + f = 0

は2次曲線とも呼ばれ、

\begin{array}{r} (\begin{array}{ll} x & y \end{array}) (\begin{array}{ll} a & b \\ b & c \end{array}) (\binom{x}{y}) + 2 (\begin{array}{ll} d & e \end{array}) (\binom{x}{y}) + f = 0 \end{array}

と表される。ここで、行列 $(\begin{array}{ll} a & b \\ b & c \end{array})$ は対称行列であることに注意しよう。対称行列の直交行列による対角化を用いると、上記のような方程式を標準形という、よりわかりやすい式で表すことができる。例えば、楕円、双曲線、放物線の標準形はそれぞれ

\frac{x^{2}}{a^{2}} + \frac{y^{2}}{b^{2}} = 1, \frac{x^{2}}{a^{2}} - \frac{y^{2}}{b^{2}} = 1, y = a x^{2}

で表される。ただし、 $a, b$ は 0 ではない定数である。

問2#

直交行列の固有値は絶対値 1 の複素数であることを示せ。

$A$ を $n$ 次の直交行列、 $x$ をゼロでない $n$ 次のベクトルとする。

A x = λ x

のノルムを取ると

\begin{array}{r} \begin{aligned} ‖ A x ‖ & = ‖ λ x ‖ \\ ‖ A x ‖ & = | λ | ‖ x ‖ (ノ ル ム の 定 数 倍 に つ い て の 定 理 ‖ c x ‖ = | c | ‖ x ‖ よ り) \\ ‖ x ‖ & = | λ | ‖ x ‖ (直 交 変 換 は 長 さ を 保 つ 、 す な わ ち ‖ A x ‖ = ‖ x ‖ の た め) \end{aligned} \end{array}

よって $| λ | = 1$ となる。

固有値は一般には複素数となるため、直交行列の固有値は絶対値1の複素数である

練習問題メモ 24（対称行列の対角化）

Contents

練習問題メモ 24（対称行列の対角化）#

問1#

問2#