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, 祭音Myyura

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)、$\lim_{n \rightarrow \infty} \left (\begin{array}{cccc} x_{n} \\ y_{n} \\ z_{n} \\ \end{array}\right)$ 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．

题目描述

三维向量满足递推

$$\begin{pmatrix}x_{n+1}\\y_{n+1}\\z_{n+1}\end{pmatrix} =A\begin{pmatrix}x_n\\y_n\\z_n\end{pmatrix}, \qquad n=0,1,2,\ldots,$$

其中 $x_0,y_0,z_0,\alpha$ 为实数，

$$A=\begin{pmatrix} 1-2\alpha&\alpha&\alpha\\ \alpha&1-\alpha&0\\ \alpha&0&1-\alpha \end{pmatrix}, \qquad0<\alpha<\frac13.$$

回答下列问题。

（1）用 $x_0,y_0,z_0$ 表示 $x_n+y_n+z_n$。

（2）求 $A$ 的三个特征值 $\lambda_1,\lambda_2,\lambda_3$ 及对应特征向量 $v_1,v_2,v_3$。

（3）用这些特征值和特征向量表示矩阵 $A$。

（4）用 $x_0,y_0,z_0,\alpha$ 显式表示 $(x_n,y_n,z_n)^{\mathsf T}$。

（5）求

$$\lim_{n\to\infty} \begin{pmatrix}x_n\\y_n\\z_n\end{pmatrix}.$$

（6）在 $x_0^2+y_0^2+z_0^2\ne0$ 下，把

$$f(x_0,y_0,z_0)= \frac{(x_0,y_0,z_0) \begin{pmatrix}x_{n+1}\\y_{n+1}\\z_{n+1}\end{pmatrix}} {(x_0,y_0,z_0) \begin{pmatrix}x_n\\y_n\\z_n\end{pmatrix}}$$

视为初始向量的函数，求其最大值和最小值。

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 $A$ is symmetric and $q_1,q_2,q_3$ are orthonormal,

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

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,

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

Therefore,

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

(6)

Let $p_0=(x_0,y_0,z_0)^T$ and $c_i=q_i^Tp_0$. Since $p_n=A^np_0$,

$$f(x_0,y_0,z_0) =\frac{p_0^TA^{n+1}p_0}{p_0^TA^np_0} =\frac{\sum_{i=1}^3\lambda_i^{n+1}c_i^2} {\sum_{i=1}^3\lambda_i^nc_i^2}.$$

All $\lambda_i$ are positive, so the denominator is positive and this is a weighted average of $\lambda_1,\lambda_2,\lambda_3$. Therefore,

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

attained when $p_0$ is parallel to $q_1$, while

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

is attained when $p_0$ is parallel to $q_3$.