早稲田大学 創造理工学研究科 経営システム工学専攻 2014年8月実施 线性代数

标签：

• 早稻田大学
• 特征值与特征向量

Author

思齐塾, 祭音Myyura

Description

次の行列(matrix)の固有値 (eigenvalue) と固有ベクトル (eigenvector) をすべて求めよ。

$$\begin{bmatrix} -1 & 0 & 0 \\ 1 & 2 & 0 \\ 3 & -4 & 1 \end{bmatrix}$$

题目描述

求矩阵

$$\begin{bmatrix} -1&0&0\\ 1&2&0\\ 3&-4&1 \end{bmatrix}$$

的全部特征值及其对应的全部特征向量。

Kai

Let

$$A = \begin{bmatrix} -1 & 0 & 0 \\ 1 & 2 & 0 \\ 3 & -4 & 1 \end{bmatrix}$$

.

To find the eigenvalues, we need to solve the characteristic equation $|A - \lambda I| = 0$ , where $I$ is the identity matrix.

$$A - \lambda I = \begin{bmatrix} -1 - \lambda & 0 & 0 \\ 1 & 2 - \lambda & 0 \\ 3 & -4 & 1 - \lambda \end{bmatrix}$$

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

So, the eigenvalues are $\lambda_1 = -1$ , $\lambda_2 = 2$ , and $\lambda_3 = 1$ .

Now, let's find the eigenvectors for each eigenvalue.

For $\lambda_1 = -1$ : $(A - \lambda_1 I)v_1 = 0$

$$\begin{bmatrix} 0 & 0 & 0 \\ 1 & 3 & 0 \\ 3 & -4 & 2 \end{bmatrix} \begin{bmatrix} x_1 \\ x_2 \\ x_3 \end{bmatrix} = \begin{bmatrix} 0 \\ 0 \\ 0 \end{bmatrix}$$

$x_1 + 3x_2 = 0 \Rightarrow x_1 = -3x_2$ $3x_1 - 4x_2 + 2x_3 = 0 \Rightarrow 3(-3x_2) - 4x_2 + 2x_3 = 0 \Rightarrow -13x_2 + 2x_3 = 0 \Rightarrow x_3 = \frac{13}{2}x_2$

Let $x_2 = 2$ , then $x_1 = -6$ and $x_3 = 13$ . So,

$$v_1 = \begin{bmatrix} -6 \\ 2 \\ 13 \end{bmatrix}$$

For $\lambda_2 = 2$ : $(A - \lambda_2 I)v_2 = 0$

$$\begin{bmatrix} -3 & 0 & 0 \\ 1 & 0 & 0 \\ 3 & -4 & -1 \end{bmatrix} \begin{bmatrix} x_1 \\ x_2 \\ x_3 \end{bmatrix} = \begin{bmatrix} 0 \\ 0 \\ 0 \end{bmatrix}$$

$-3x_1 = 0 \Rightarrow x_1 = 0$ $x_1 = 0$ $3x_1 - 4x_2 - x_3 = 0 \Rightarrow -4x_2 - x_3 = 0 \Rightarrow x_3 = -4x_2$

Let $x_2 = 1$ , then $x_3 = -4$ . So,

$$v_2 = \begin{bmatrix} 0 \\ 1 \\ -4 \end{bmatrix}$$

For $\lambda_3 = 1$ : $(A - \lambda_3 I)v_3 = 0$

$$\begin{bmatrix} -2 & 0 & 0 \\ 1 & 1 & 0 \\ 3 & -4 & 0 \end{bmatrix} \begin{bmatrix} x_1 \\ x_2 \\ x_3 \end{bmatrix} = \begin{bmatrix} 0 \\ 0 \\ 0 \end{bmatrix}$$

$-2x_1 = 0 \Rightarrow x_1 = 0$ $x_1 + x_2 = 0 \Rightarrow x_2 = 0$ $3x_1 - 4x_2 = 0 \Rightarrow 0 = 0$

Let $x_3 = 1$ . So,

$$v_3 = \begin{bmatrix} 0 \\ 0 \\ 1 \end{bmatrix}$$

Therefore, the eigenvalues are -1, 2, and 1, with corresponding eigenvectors

$$\begin{bmatrix} -6 \\ 2 \\ 13 \end{bmatrix}$$

,

$$\begin{bmatrix} 0 \\ 1 \\ -4 \end{bmatrix}$$

, and

$$\begin{bmatrix} 0 \\ 0 \\ 1 \end{bmatrix}$$

.

For each eigenvalue, all eigenvectors are the nonzero scalar multiples of its displayed vector.