広島大学 先進理工系科学研究科 情報科学プログラム 2019年8月実施 専門科目I 問題1

标签：

• Hiroshima-University
• Eigenvalues-and-Eigenvectors

Author

samparker, 祭音Myyura

Description

数列

$$\begin{aligned} &a_0 = 0, a_1 = 1 \\ &a_{n+2} = a_{n+1} + a_n \quad (n = 0, 1, 2, \ldots) \end{aligned}$$

の漸化式は、行列 $A = \begin{bmatrix} 1 & 1 \\ 1 & 0 \end{bmatrix}$ を用いて、

$$\begin{bmatrix} a_{n+2} \\ a_{n+1} \end{bmatrix} = \begin{bmatrix} 1 & 1 \\ 1 & 0 \end{bmatrix} \begin{bmatrix} a_{n+1} \\ a_{n} \end{bmatrix}$$

と表現することができる。以下の問題を答えよ。

(1) 行列 $A$ のすべての固有値と固有ベクトルを求めよ。

(2) この数列の一般項 $a_n$ を求めよ。

---

Let

$$\begin{aligned} &a_0 = 0, a_1 = 1 \\ &a_{n+2} = a_{n+1} + a_n \quad (n = 0, 1, 2, \ldots) \end{aligned}$$

be a sequence of numbers. The recurrence relation of this sequence can be represented by using matrix $A = \begin{bmatrix} 1 & 1 \\ 1 & 0 \end{bmatrix}$ as

$$\begin{bmatrix} a_{n+2} \\ a_{n+1} \end{bmatrix} = \begin{bmatrix} 1 & 1 \\ 1 & 0 \end{bmatrix} \begin{bmatrix} a_{n+1} \\ a_{n} \end{bmatrix}$$

(1) Find all the eigenvalues and the corresponding eigenvectors of $A$.

(2) Find a general term $a_n$ of this sequence.

题目描述

数列满足

$$a_0=0,\qquad a_1=1,\qquad a_{n+2}=a_{n+1}+a_n\quad(n=0,1,2,\ldots),$$

其递推关系可借助

$$A=\begin{bmatrix}1&1\\1&0\end{bmatrix}$$

写成

$$\begin{bmatrix}a_{n+2}\\a_{n+1}\end{bmatrix} =A\begin{bmatrix}a_{n+1}\\a_n\end{bmatrix}.$$

1. 求矩阵 $A$ 的全部特征值及对应的特征向量。
2. 求该数列的通项 $a_n$。

考点

• 特征值与特征向量：对递推矩阵作谱分解，并利用矩阵幂推导斐波那契型数列的通项公式。

Kai

(1)

Eigenvalues

$$\text{det}(A - \lambda E) = 0 \Rightarrow \begin{vmatrix} 1 - \lambda & 1 \\ 1 & -\lambda \end{vmatrix} = \lambda^2 - \lambda - 1 = 0$$

$$\therefore \lambda_1 = \frac{1 + \sqrt{5}}{2}, \lambda_2 = \frac{1 - \sqrt{5}}{2}$$

Eigenvectors

$$v_1 = \left(\frac{1 + \sqrt{5}}{2}, 1 \right), v_2 = \left(\frac{1 - \sqrt{5}}{2}, 1 \right)$$

(2)

Let $P$ be the matrix formed by these eigenvectors, i.e.

$$P = \begin{bmatrix} \frac{1 + \sqrt{5}}{2} & \frac{1 - \sqrt{5}}{2} \\ 1 & 1 \end{bmatrix}$$

Then

$$\text{det}(P) = \lambda_1 - \lambda_2 = \sqrt{5}$$

In particular,

$$P^{-1} = \frac{1}{\sqrt{5}} \begin{bmatrix} 1 & -\lambda_2 \\ -1 & \lambda_1 \end{bmatrix}$$

Then the diagonalization of $A$ is given by

$$A = \begin{bmatrix} \lambda_1 & \lambda_2 \\ 1 & 1 \end{bmatrix} \begin{bmatrix} \lambda_1 & 0 \\ 0 & \lambda_2 \end{bmatrix} \frac{1}{\sqrt{5}} \begin{bmatrix} 1 & -\lambda_2 \\ -1 & \lambda_1 \end{bmatrix}$$

The $n$th-power of $A$ is

$$\begin{aligned} A^n &= \begin{bmatrix} \lambda_1 & \lambda_2 \\ 1 & 1 \end{bmatrix} \begin{bmatrix} \lambda_1^n & 0 \\ 0 & \lambda_2^n \end{bmatrix} \frac{1}{\sqrt{5}} \begin{bmatrix} 1 & -\lambda_2 \\ -1 & \lambda_1 \end{bmatrix} \\ &= \begin{bmatrix} \lambda_1^{n+1} & \lambda_2^{n+1} \\ \lambda_1^n & \lambda_2^n \end{bmatrix} \frac{1}{\sqrt{5}} \begin{bmatrix} 1 & -\lambda_2 \\ -1 & \lambda_1 \end{bmatrix} \end{aligned}$$

Then the $(1, 2)$-entry of $A_n$ is

$$\begin{aligned} a_n &= \frac{1}{\sqrt{5}} (-\lambda_1^{n+1} \lambda_2 + \lambda_2^{n+1} \lambda_1) \\ &= \frac{1}{\sqrt{5}} \lambda_1 \lambda_2 (\lambda_2^n - \lambda_1^n) \\ &= \frac{1}{\sqrt{5}} (\lambda_1^n - \lambda_2^n) \\ &= \frac{1}{\sqrt{5}} \left( \left( \frac{1 + \sqrt{5}}{2} \right)^n - \left( \frac{1 - \sqrt{5}}{2} \right)^n \right) \end{aligned}$$