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

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.

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

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