早稲田大学 創造理工学研究科 経営システム工学専攻 2013年8月実施 线性代数

标签：

• Waseda-University
• Eigenvalues-and-Eigenvectors

Author

思齐塾, 祭音Myyura

Description

次の行列の固有値と固有ベクトルをすべて求めよ.

$$\begin{pmatrix} 1 & 1 & 0 & 0 \\ 0 & 2 & 1 & 0 \\ 0 & 0 & 2 & 1 \\ 0 & 0 & 0 & 1 \end{pmatrix}$$

题目描述

求矩阵

$$\begin{pmatrix} 1&1&0&0\\ 0&2&1&0\\ 0&0&2&1\\ 0&0&0&1 \end{pmatrix}$$

的全部特征值及其对应的全部特征向量。

Kai

Let $A = \begin{pmatrix} 1 & 1 & 0 & 0 \\ 0 & 2 & 1 & 0 \\ 0 & 0 & 2 & 1 \\ 0 & 0 & 0 & 1 \end{pmatrix}$ . To find the eigenvalues, we solve the characteristic equation $\det(A - \lambda I) = 0$ .

$$A - \lambda I = \begin{pmatrix} 1-\lambda & 1 & 0 & 0 \\ 0 & 2-\lambda & 1 & 0 \\ 0 & 0 & 2-\lambda & 1 \\ 0 & 0 & 0 & 1-\lambda \end{pmatrix}$$

$$\det(A - \lambda I) = (1-\lambda)(2-\lambda)(2-\lambda)(1-\lambda) = (1-\lambda)^2(2-\lambda)^2 = 0$$

So the eigenvalues are $\lambda_1 = 1$ (with multiplicity 2) and $\lambda_2 = 2$ (with multiplicity 2).

For $\lambda_1 = 1$ :

$$A - I = \begin{pmatrix} 0 & 1 & 0 & 0 \\ 0 & 1 & 1 & 0 \\ 0 & 0 & 1 & 1 \\ 0 & 0 & 0 & 0 \end{pmatrix}$$

Solving $(A - I)v = 0$ , we get: $v_2 = 0$ $v_2 + v_3 = 0$ , so $v_3 = 0$ $v_3 + v_4 = 0$ , so $v_4 = 0$ Thus $v = \begin{pmatrix} v_1 \\ 0 \\ 0 \\ 0 \end{pmatrix} = v_1 \begin{pmatrix} 1 \\ 0 \\ 0 \\ 0 \end{pmatrix}$ . Eigenvector for $\lambda_1 = 1$ is $v_1 = \begin{pmatrix} 1 \\ 0 \\ 0 \\ 0 \end{pmatrix}$ .

Let's find a generalized eigenvector. We need to solve $(A-I)^2 w = 0$ . We want to find $w$ such that $(A-I)w = v_1$ .

$$(A - I) w = \begin{pmatrix} 0 & 1 & 0 & 0 \\ 0 & 1 & 1 & 0 \\ 0 & 0 & 1 & 1 \\ 0 & 0 & 0 & 0 \end{pmatrix} \begin{pmatrix} w_1 \\ w_2 \\ w_3 \\ w_4 \end{pmatrix} = \begin{pmatrix} 1 \\ 0 \\ 0 \\ 0 \end{pmatrix}$$

$w_2 = 1$ $w_2 + w_3 = 0$ , so $w_3 = -1$ $w_3 + w_4 = 0$ , so $w_4 = 1$ Thus $w = \begin{pmatrix} w_1 \\ 1 \\ -1 \\ 1 \end{pmatrix}$ . Choosing $w_1 = 0$ , we get $w = \begin{pmatrix} 0 \\ 1 \\ -1 \\ 1 \end{pmatrix}$ . This is a generalized eigenvector.

For $\lambda_2 = 2$ :

$$A - 2I = \begin{pmatrix} -1 & 1 & 0 & 0 \\ 0 & 0 & 1 & 0 \\ 0 & 0 & 0 & 1 \\ 0 & 0 & 0 & -1 \end{pmatrix}$$

Solving $(A - 2I)v = 0$ , we get: $-v_1 + v_2 = 0$ , so $v_1 = v_2$ $v_3 = 0$ $v_4 = 0$ $-v_4 = 0$ Thus $v = \begin{pmatrix} v_2 \\ v_2 \\ 0 \\ 0 \end{pmatrix} = v_2 \begin{pmatrix} 1 \\ 1 \\ 0 \\ 0 \end{pmatrix}$ . Eigenvector for $\lambda_2 = 2$ is $v_2 = \begin{pmatrix} 1 \\ 1 \\ 0 \\ 0 \end{pmatrix}$ .

Let's find a generalized eigenvector. We need to solve $(A-2I)^2 w = 0$ . We want to find $w$ such that $(A-2I)w = v_2$ .

$$(A - 2I) w = \begin{pmatrix} -1 & 1 & 0 & 0 \\ 0 & 0 & 1 & 0 \\ 0 & 0 & 0 & 1 \\ 0 & 0 & 0 & -1 \end{pmatrix} \begin{pmatrix} w_1 \\ w_2 \\ w_3 \\ w_4 \end{pmatrix} = \begin{pmatrix} 1 \\ 1 \\ 0 \\ 0 \end{pmatrix}$$

$-w_1 + w_2 = 1$ $w_3 = 1$ $w_4 = 0$ $-w_4 = 0$ Thus $w = \begin{pmatrix} w_1 \\ w_1 + 1 \\ 1 \\ 0 \end{pmatrix}$ . Choosing $w_1 = 0$ , we have $w = \begin{pmatrix} 0 \\ 1 \\ 1 \\ 0 \end{pmatrix}$ .

Therefore, the eigenvalues are $1$ and $2$ , each with algebraic multiplicity $2$ , and all corresponding eigenvectors are

$$\ker(A-I)=\operatorname{span}\left\{ \begin{pmatrix}1\\0\\0\\0\end{pmatrix} \right\}, \qquad \ker(A-2I)=\operatorname{span}\left\{ \begin{pmatrix}1\\1\\0\\0\end{pmatrix} \right\}.$$

All eigenvectors are the nonzero vectors in these two eigenspaces. The vectors $\begin{pmatrix}0\\1\\-1\\1\end{pmatrix}$ and $\begin{pmatrix}0\\1\\1\\0\end{pmatrix}$ found above are generalized eigenvectors, not eigenvectors.