特殊矩陣 (11)：三對角矩陣

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

$A^{-1}=\left[\!\!\begin{array}{rrrr} 1.750&-0.250&-1.500&0.750\\ -0.375&0.125&0.750&-0.375\\ -1.500&0.500&1.000&-0.500\\ 2.250&-0.750&-1.500&1.250 \end{array}\!\!\right]$

$A_3=\begin{bmatrix} a_{1}&b_{1}&0\\ c_{1}&a_2&b_2\\ 0&c_2&a_3 \end{bmatrix}$

$\det A_3=a_3\begin{vmatrix} a_1&b_1\\ c_1&a_2 \end{vmatrix}-c_2\begin{vmatrix} a_1&0\\ c_1&b_2 \end{vmatrix}=a_3\det A_2-b_2c_2\det A_1$

$\det A_{k+1}=a_{k+1}\det A_{k}-b_{k}c_{k}\det A_{k-1}$

\begin{aligned} \det A_1&=1\\ \det A_2&=-4\\ \det A_3&=1\cdot\det A_2-3\cdot 2\cdot\det A_1=-10\\ \det A_4&=2\cdot\det A_3-1\cdot 3\cdot\det A_2=-8\end{aligned}

$D^{-1}A_3D=\begin{bmatrix} 1/d_1&0&0\\ 0&1/d_2&0\\ 0&0&1/d_3 \end{bmatrix}\begin{bmatrix} a_1&b_1&0\\ c_1&a_2&b_2\\ 0&c_2&a_3 \end{bmatrix}\begin{bmatrix} d_1&0&0\\ 0&d_2&0\\ 0&0&d_3 \end{bmatrix}$

$D^{-1}A_3D=\begin{bmatrix} a_1&b_1d_2/d_1&0\\ c_1d_1/d_2&a_2&b_2d_3/d_2\\ 0&c_2d_2/d_3&a_3 \end{bmatrix}$

$\displaystyle \frac{b_1d_2}{d_1}=\frac{c_1d_1}{d_2},~\frac{b_2d_3}{d_2}=\frac{c_2d_2}{d_3}$

$\displaystyle \frac{b_id_{i+1}}{d_i}=\displaystyle{c_id_i}{d_{i+1}}$

$\displaystyle d_{i+1}^2=\left(\frac{c_i}{b_i}\right)d_i^2$

$d_{i+1}=\displaystyle\sqrt{\frac{c_i}{b_i}}\cdot d_i$

$\begin{bmatrix} a_{1}&b_{1}&0\\ c_{1}&a_2&b_2\\ 0&c_2&a_3 \end{bmatrix}\begin{bmatrix} x_1\\x_2\\x_3 \end{bmatrix}=\begin{bmatrix} d_1\\d_2\\d_3 \end{bmatrix}$

$\displaystyle b_1^{\prime}=\frac{b_1}{a_1},~~d_1^{\prime}=\frac{d_1}{a_1}$

$\begin{bmatrix} 1&b_{1}^{\prime}&0\\ c_{1}&a_2&b_2\\ 0&c_2&a_3 \end{bmatrix}\begin{bmatrix} x_1\\x_2\\x_3 \end{bmatrix}=\begin{bmatrix} d_1^{\prime}\\d_2\\d_3 \end{bmatrix}$

$\displaystyle b'_2=\frac{b_2}{a_2-b'_1c_1},~~ d'_2=\frac{d_2-d_1^{\prime}c_1}{a_2-b'_1c_1}$

$\begin{bmatrix} 1&b_{1}^{\prime}&0\\ 0&1&b_2^{\prime}\\ 0&c_2&a_3 \end{bmatrix}\begin{bmatrix} x_1\\x_2\\x_3 \end{bmatrix}=\begin{bmatrix} d_1^{\prime}\\d_2^{\prime}\\d_3 \end{bmatrix}$

$\displaystyle d_3^{\prime}=\frac{d_3-d_2^{\prime}c_2}{a_3-b_2^{\prime}c_2}$

$\begin{bmatrix} 1&b_{1}^{\prime}&0\\ 0&1&b_2^{\prime}\\ 0&0&1 \end{bmatrix}\begin{bmatrix} x_1\\x_2\\x_3 \end{bmatrix}=\begin{bmatrix} d_1^{\prime}\\d_2^{\prime}\\d_3^{\prime} \end{bmatrix}$

\begin{aligned} x_3&=d_3^{\prime}\\ x_2&=d_2^{\prime}-b_2^{\prime}x_3\\ x_1&=d_1^{\prime}-b_1^{\prime}x_2\end{aligned}

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4 Responses to 特殊矩陣 (11)：三對角矩陣

1. Anonymous says:

請問關於三對角矩陣的特徵值有無相關文章或公式？

2. ccjou says:

三對角矩陣的特徵值沒有代數公式解，特殊形式矩陣的特徵方程可化為二階差分方程，由此可解出特徵值，請見例子
https://ccjou.wordpress.com/2010/02/10/%E6%AF%8F%E9%80%B1%E5%95%8F%E9%A1%8C-february-15-2010/

QR 演算法是一個相當有效的三對角矩陣特徵值數值計算法，請見
Numerical Recipes in FORTRAN 77: The Art of Scientific Computing

Click to access f11-3.pdf

3. CCH says:

內容文字第9行的–A3的”主子陣”==>應該是”領先主子陣”?

• ccjou says:

太對了，已更正。感謝。