東京大学 情報理工学研究科 2017年度 数学 第1問

Author

Zero, etsurin

Description

3次元ベクトル \(\left (\begin{array}{cccc} x_{n} \\ y_{n} \\ z_{n} \\ \end{array}\right)\) は式

\[ \left (\begin{array}{cccc} x_{n+1} \\ y_{n+1} \\ z_{n+1} \\ \end{array}\right)=A \left (\begin{array}{cccc} x_{n} \\ y_{n} \\ z_{n} \\ \end{array}\right) (n=0,1,2,...) \]

を満たすものとする．ただし，\(x_{0},y_{0},z_{0},\alpha\) は実数とし，

\[ A=\left (\begin{array}{cccc} 1-2\alpha & \alpha &\alpha \\ \alpha &1-\alpha &0 \\ \alpha &0 &1-\alpha \\ \end{array}\right),0<\alpha<\frac{1}{3} \]

とする．以下の問いに答えよ.

(1)、\(x_{n},y_{n},z_{n}\) を \(x_{0},y_{0},z_{0}\) を用いて表せ．

(2)、行列 \(A\) の固有値 \(\lambda_1,\lambda_2,\lambda_3\) と，それぞれの固有値に対応する固有ベクトル \(v_{1},v_{2},v_{3}\) を求めよ．

(3)、行列 \(A\) を \(\lambda_1,\lambda_2,\lambda_3,v_{1},v_{2},v_{3}\) を用いて表せ．

(4)、\(\left (\begin{array}{cccc} x_{n} \\ y_{n} \\ z_{n} \\ \end{array}\right)\) を \(x_{0},y_{0},z_{0},\alpha\) を用いて表せ．

(5)、\(\lim_{x \rightarrow \infty} \left (\begin{array}{cccc} x_{n} \\ y_{n} \\ z_{n} \\ \end{array}\right)\) を求めよ．

(6)、以下の式

\[ f(x_{0},y_{0},z_{0})=\frac{(x_{0},y_{0},z_{0}) \left (\begin{array}{cccc} x_{n+1} \\ y_{n+1} \\ z_{n+1} \\ \end{array}\right)} {(x_{0},y_{0},z_{0}) \left (\begin{array}{cccc} x_{n} \\ y_{n} \\ z_{n} \\ \end{array}\right)} \]

を \(x_{0},y_{0},z_{0}\) の関数とみなして，\(f(x_{0},y_{0},z_{0})\) の最大値および最小値お求めよ．ただし，\(x_{0}^2+y_{0}^2+z_{0}^2 \neq 0\) とする．

Kai

(1)

By the following given equations:

\[ \left (\begin{array}{cccc} x_{n+1} \\ y_{n+1} \\ z_{n+1} \\ \end{array}\right) = A \left (\begin{array}{cccc} x_{n} \\ y_{n} \\ z_{n} \\ \end{array}\right) = \left (\begin{array}{cccc} 1-2\alpha &\alpha & \alpha \\ \alpha & 1-\alpha & 0 \\ \alpha & 0 & 1-\alpha \\ \end{array}\right) \left (\begin{array}{cccc} x_{n} \\ y_{n} \\ z_{n} \\ \end{array}\right) \]

We have

\[ x_{n+1}+y_{n+1}+z_{n+1}= (1\ 1\ 1) \left (\begin{array}{cccc} x_{n+1} \\ y_{n+1} \\ z_{n+1} \\ \end{array}\right)= (1\ 1\ 1) A \left (\begin{array}{cccc} x_{n} \\ y_{n} \\ z_{n} \\ \end{array}\right)=x_{n}+y_{n}+z_{n} \]

Therefore:

\[ x_{n}+y_{n}+z_{n} = x_{n-1}+y_{n-1}+z_{n-1} = \cdots = x_{0}+y_{0}+z_{0} \]

(2)

\[ \begin{aligned} \det (A - \lambda I) &= \begin{vmatrix} 1 - 2\alpha - \lambda & \alpha & \alpha \\ \alpha & 1 - \alpha - \lambda & 0 \\ \alpha & 0 & 1 - \alpha - \lambda \end{vmatrix} \\ &= (\lambda^2 + (3\alpha - 2)\lambda + (1-3\alpha))(1 - \alpha - \lambda) \\ &= 0 \end{aligned} \]

Hence,

\[ \lambda_{1}=1, v_{1}= \begin{pmatrix} 1 \\ 1 \\ 1 \end{pmatrix} \]

\[ \lambda_{2}= 1-\alpha, v_{2}= \begin{pmatrix} 0 \\ 1 \\ -1 \end{pmatrix} \]

\[ \lambda_{3}= 1-3\alpha, v_{2}= \begin{pmatrix} -2 \\ 1 \\ 1 \end{pmatrix} \]

(3)

From (2), we know that

\[ A(v_{1}\ v_{2}\ v_{3})=(v_{1}\ v_{2}\ v_{3})\text{diag}(\lambda_{1}\ \lambda_{2}\ \lambda_{3}) \]

Hence,

\[ A=(v_{1}\ v_{2}\ v_{3})\text{diag}(\lambda_{1}\ \lambda_{2}\ \lambda_{3})(v_{1}\ v_{2}\ v_{3})^{-1} \]

Which is,

\[ A=\left (\begin{array}{cccc} 1 &0 & -2\\ 1 &-1 & 1\\ 1 &1 & 1\\ \end{array}\right) \left (\begin{array}{cccc} 1 &0 & 0\\ 1 &1-\alpha &0\\ 0 &0 &1-3\alpha\\ \end{array}\right) \left (\begin{array}{cccc} 1 &0 & -2\\ 1 &-1 & 1\\ 1 &1 & 1\\ \end{array}\right)^{-1} \]

(4)

Note that,

\[ \begin{aligned} \left (\begin{array}{cccc} x_{n} \\ y_{n} \\ z_{n} \\ \end{array}\right)&=A^{n} \left (\begin{array}{cccc} x_{0} \\ y_{0} \\ z_{0} \\ \end{array}\right) \\ &= ((v_{1}\ v_{2}\ v_{3})\text{diag}(\lambda_{1}\ \lambda_{2}\ \lambda_{3})(v_{1}\ v_{2}\ v_{3})^{-1})^{n} \left (\begin{array}{cccc} x_{0} \\ y_{0} \\ z_{0} \\ \end{array}\right) \\ &= (v_{1}\ v_{2}\ v_{3})\text{diag}(\lambda_{1}\ \lambda_{2}\ \lambda_{3})^{n}(v_{1}\ v_{2}\ v_{3})^{-1} \left (\begin{array}{cccc} x_{0} \\ y_{0} \\ z_{0} \\ \end{array}\right) \end{aligned} \]

Normalize the characteristic vectors:

\[ q_{1}=\frac{v_{1}}{\Vert v_{1}\Vert}= \begin{pmatrix} \frac{1}{\sqrt{3}} \\ \frac{1}{\sqrt{3}} \\ \frac{1}{\sqrt{3}} \end{pmatrix} \]

\[ q_{2}=\frac{v_{2}}{\Vert v_{2}\Vert}= \begin{pmatrix} 0 \\ \frac{1}{\sqrt{2}} \\ -\frac{1}{\sqrt{2}} \end{pmatrix} \]

\[ q_{3}=\frac{v_{3}}{\Vert v_{3}\Vert} = \begin{pmatrix} -\frac{2}{\sqrt{6}} \\ \frac{1}{\sqrt{6}} \\ \frac{1}{\sqrt{6}} \end{pmatrix} \]

We have

\[ \left (\begin{array}{cccc} x_{n} \\ y_{n} \\ z_{n} \\ \end{array}\right)= (q_{1}\ q_{2}\ q_{3})\text{diag}(\lambda_{1}^{n},\lambda_{2}^{n},\lambda_{3}^{n})(q_{1}\ q_{2}\ q_{3})^{-1} \left (\begin{array}{cccc} x_{n} \\ y_{n} \\ z_{n} \\ \end{array}\right) \]

Since \(0<\alpha<\frac{1}{3}\), \(A\) is positive-definite.

\[ (q_{1}\ q_{2}\ q_{3})^{-1}= (q_{1}\ q_{2}\ q_{3})^{T}= \left (\begin{array}{cccc} q_{1}^{T} \\ q_{2}^{T} \\ q_{3}^{T} \\ \end{array}\right) \]

Hence,

\[ \left (\begin{array}{cccc} x_{n} \\ y_{n} \\ z_{n} \\ \end{array}\right)= (\lambda_{1}^{n}q_{1}q_{1}^{T}+ \lambda_{2}^{n}q_{2}q_{2}^{T}+ \lambda_{3}^{n}q_{3}q_{3}^{T}) \left (\begin{array}{cccc} x_{0} \\ y_{0} \\ z_{0} \\ \end{array}\right) \]

where

\[ q_{1}q_{1}^{T}= \left (\begin{array}{cccc} \frac{1}{3} &\frac{1}{3} &\frac{1}{3}\\ \frac{1}{3} &\frac{1}{3} &\frac{1}{3}\\ \frac{1}{3} &\frac{1}{3} &\frac{1}{3}\\ \end{array}\right), \]

\[ q_{2}q_{2}^{T}= \left (\begin{array}{cccc} 0 &0 &0\\ 0 &\frac{1}{2} &-\frac{1}{2}\\ 0 &-\frac{1}{2} &\frac{1}{2}\\ \end{array}\right), \]

\[ q_{3}q_{3}^{T}= \left (\begin{array}{cccc} \frac{2}{3} &-\frac{1}{3} &-\frac{1}{3}\\ -\frac{1}{3} &\frac{1}{6} &\frac{1}{6}\\ -\frac{1}{3} &\frac{1}{6} &\frac{1}{6}\\ \end{array}\right) \]

(5)

Since

\[ \vert\lambda_{1}\vert=1, \vert\lambda_{2}\vert=1-\alpha<1, \vert\lambda_{3}\vert=1-3\alpha<1 \]

Hence we have,

\[ \begin{aligned} \lim_{x \rightarrow \infty} \left (\begin{array}{cccc} x_{n} \\ y_{n} \\ z_{n} \\ \end{array}\right)&= (\lim_{x \rightarrow \infty}\lambda_{1}^{n}q_{1}q_{1}^{T}+\lim_{x \rightarrow \infty}\lambda_{2}^{n}q_{2}q_{2}^{T}+\lim_{x \rightarrow \infty}\lambda_{3}^{n}q_{3}q_{3}^{T}) \left (\begin{array}{cccc} x_{0} \\ y_{0} \\ z_{0} \\ \end{array}\right) \\ &= q_{1}q_{1}^{T} \left (\begin{array}{cccc} x_{0} \\ y_{0} \\ z_{0} \\ \end{array}\right) \end{aligned} \]

Therefore,

\[ \lim_{x \rightarrow \infty} \left (\begin{array}{cccc} x_{n} \\ y_{n} \\ z_{n} \\ \end{array}\right)= \frac{1}{3}(x_{0}+y_{0}+z_{0}) \left (\begin{array}{cccc} 1 \\ 1 \\ 1 \\ \end{array}\right) \]

(6)

Note that

\[ f(x_{0},y_{0},z_{0})= \frac{\begin{pmatrix} x_{n} & y_{n} & z_{n} \end{pmatrix} A \left (\begin{array}{cccc} x_{n} \\ y_{n} \\ z_{n} \\ \end{array}\right)} {\begin{pmatrix} x_{n} & y_{n} & z_{n} \end{pmatrix} \left (\begin{array}{cccc} x_{n} \\ y_{n} \\ z_{n} \\ \end{array}\right)} \]

Let

\[ p_{n}=\left (\begin{array}{cccc} x_{n} \\ y_{n} \\ z_{n} \\ \end{array}\right) \]

where

\[ \begin{pmatrix} x_{n} & y_{n} & z_{n} \end{pmatrix} \left (\begin{array}{cccc} x_{n} \\ y_{n} \\ z_{n} \\ \end{array}\right)= x_{n}^{2}+y_{n}^{2}+z_{n}^{2}=\Vert p_{n} \Vert ^{2} \]

and

\[ \begin{pmatrix} x_{n} & y_{n} & z_{n} \end{pmatrix} A \left (\begin{array}{cccc} x_{n} \\ y_{n} \\ z_{n} \\ \end{array}\right)= \lambda_{1}p_{n}^{T}(q_{1}q_{1}^{T}p_{n})+\lambda_{2}p_{n}^{T}(q_{2}q_{2}^{T}p_{n})+\lambda_{3}p_{n}^{T}(q_{3}q_{3}^{T}p_{n}) \]

Since A is positive-definite, \(q_{1},q_{2},q_{3}\) are all orthogonal, that is:

\[ q_{1}\perp q_{2},q_{2}\perp q_{3},q_{3}\perp q_{1} \]

if \(p_{n}//q_{1}\), then \(p_{n}\perp q_{2}\) and \(p_{n}\perp q_{3}\), the maximum of \(f(x_{0},y_{0},z_{0})\) is

\[ \max(f(x_{0},y_{0},z_{0}))=\lambda_{1}=1 \]

if \(p_{n}//q_{3}\),then \(p_{n}\perp q_{1}\) and \(p_{n}\perp q_{2}\), the minimum of \(f(x_{0},y_{0},z_{0})\) is

\[ \min(f(x_{0},y_{0},z_{0}))=\lambda_{3}=1-3\alpha \]