## 利用高斯消去法計算特徵值與特徵向量

$A=\left[\!\!\begin{array}{rrr} 2& 1& 1\\ 3& -1& 4\\ -3& 2& -3 \end{array}\!\!\right]$

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

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

$\begin{bmatrix} 3& -1-\lambda& 4\\ 3(2-\lambda)& 3& 3\\ -3& 2& -3-\lambda \end{bmatrix}\to\begin{bmatrix} 3& -1-\lambda& 4\\ 0& -\lambda^2+\lambda+5&4\lambda-5\\ 0& -\lambda+1&-\lambda+1 \end{bmatrix}$

$\begin{bmatrix} 3& -\lambda-1& 4\\ 0& 5&\lambda^2+3\lambda-5\\ 0& -\lambda+1&-\lambda+1 \end{bmatrix}$

$\begin{bmatrix} 3& -\lambda-1& 4\\ 0& 5&\lambda^2+3\lambda-5\\ 0& -5(\lambda-1)&-5(\lambda-1) \end{bmatrix}\to \begin{bmatrix} 3& -\lambda-1& 4\\ 0& 5&\lambda^2+3\lambda-5\\ 0& 0&(\lambda-1)(\lambda-2)(\lambda+5)\end{bmatrix}$

$\begin{bmatrix} 3& -\lambda-1& 4\\ 0& 5&\lambda^2+3\lambda-5\\ 0& 0&0\end{bmatrix}\begin{bmatrix} x_1\\ x_2\\ x_3 \end{bmatrix}=\begin{bmatrix} 0\\ 0\\ 0 \end{bmatrix}$

\begin{aligned} x_3&=\alpha\\ x_2&=-\displaystyle\frac{\alpha}{5}(\lambda^2+3\lambda-5)\\ x_1&=-\displaystyle\frac{\alpha}{15}(\lambda+1)(\lambda^2+3\lambda-5)-\frac{4\alpha}{3}, \end{aligned}

$\displaystyle \mathbf{x}=\alpha\left[\!\!\begin{array}{r} -\frac{6}{5}\\ [0.3em] \frac{1}{5}\\ 1 \end{array}\!\!\right]$

$\lambda=2$

$\displaystyle \mathbf{x}=\alpha\left[\!\!\begin{array}{r} -\frac{7}{3}\\ [0.3em] -1\\ 1 \end{array}\!\!\right]$

$\lambda=-5$

$\displaystyle \mathbf{x}=\alpha\left[\!\!\begin{array}{r} 0\\ -1\\ 1 \end{array}\!\!\right]$

$A=\left[\!\!\begin{array}{rrr} 5&6&6\\ -1&0&-2\\ -1&-2&0 \end{array}\!\!\right]$

\begin{aligned} \begin{bmatrix} 5-\lambda&6&6\\ -1&-\lambda&-2\\ -1&-2&-\lambda \end{bmatrix}&\to\begin{bmatrix} -1&-\lambda&-2\\ 5-\lambda&6&6\\ -1&-2&-\lambda \end{bmatrix}\\ &\to\begin{bmatrix} -1&-\lambda&-2\\ 0& \lambda^2-5\lambda+6& 2(\lambda-2)\\ 0& \lambda-2& -\lambda+2 \end{bmatrix}\\ &\to\begin{bmatrix} 1&\lambda&2\\ 0& (\lambda-2)(\lambda-3)& 2(\lambda-2)\\ 0& \lambda-2& -\lambda+2 \end{bmatrix},\end{aligned}

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

$\mathbf{x}=\alpha\left[\!\!\begin{array}{r} -2\\ 1\\ 0 \end{array}\!\!\right]+\beta\left[\!\!\begin{array}{r} -2\\ 0\\ 1 \end{array}\!\!\right]$

$\lambda\neq 2$，第二列和第三列同除以 $\lambda-2$，交換第二列和第三列，再將第二列乘以 $-\lambda+3$ 加進第三列，如下：

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

$\lambda=1$$A-I$ 的列等價矩陣是

$\left[\!\!\begin{array}{ccr} 1&1&2\\ 0&1&-1\\ 0&0&0 \end{array}\!\!\right]\to\left[\!\!\begin{array}{ccr} 1&0&3\\ 0&1&-1\\ 0&0&0 \end{array}\!\!\right]$

$\mathrm{rank}(A-I)=2$，齊次解為

$\mathbf{x}=\alpha\left[\!\!\begin{array}{r} -3\\ 1\\ 1 \end{array}\!\!\right]$

[1] David C. Lay, Linear Algebra and its Applications, fourth edition, 2012, pp 280. 原文是 “Finding the characteristic polynomial of a $3\times 3$ matrix is not easy to do with just row operations, because the variable $\lambda$ is involved.”
[2] William A. McWorter, Jr. and Leroy F. Meyers, Computing eigenvalues and eigenvectors without determinants, Mathematics Magazine, Vol. 71, No. 1, 1998, pp 24-33. 原文是 “I know it is faster, but with the determinant you don’t have to think.”

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