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