東京大学 情報理工学研究科 2016年8月実施 数学 第1問

标签：

• Tokyo-University
• Eigenvalues-and-Eigenvectors
• Matrix-Diagonalization
• Matrix-Power
• Matrix-Limit
• Rayleigh-Quotient

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,

$$\lambda_{1}=1, v_{1}= \begin{pmatrix} 1 \\ 1 \\ 1 \end{pmatrix}$$

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

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

(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,

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

(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

$$\left (\begin{array}{cccc} x_{n} \\ y_{n} \\ z_{n} \\ \end{array}\right)= (q_{1}\ q_{2}\ q_{3})\text{diag}(\lambda_{1}^{n},\lambda_{2}^{n},\lambda_{3}^{n})(q_{1}\ q_{2}\ q_{3})^{-1} \left (\begin{array}{cccc} x_{0} \\ y_{0} \\ z_{0} \\ \end{array}\right)$$

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

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