# 三角矩阵

（重定向自下三角矩阵

${\displaystyle \mathbf {A} ={\begin{bmatrix}1&2\\3&4\end{bmatrix}}}$

## 描述

${\displaystyle \mathbf {L} ={\begin{bmatrix}l_{1,1}&&\cdots &&0\\l_{2,1}&l_{2,2}&&(0)&\\l_{3,1}&l_{3,2}&\ddots &&\vdots \\\vdots &\vdots &\ddots &\ddots &\\l_{n,1}&l_{n,2}&\ldots &l_{n,n-1}&l_{n,n}\end{bmatrix}}}$

${\displaystyle \mathbf {U} ={\begin{bmatrix}u_{1,1}&u_{1,2}&u_{1,3}&\ldots &u_{1,n}\\&u_{2,2}&u_{2,3}&\ldots &u_{2,n}\\\vdots &&\ddots &\ddots &\vdots \\&(0)&&\ddots &u_{n-1,n}\\0&&\cdots &&u_{n,n}\end{bmatrix}}}$

## 特殊的三角矩阵

### 高斯矩阵

${\displaystyle \mathbf {L} _{i}={\begin{bmatrix}1&&&\dots &&&&0\\0&\ddots &&&&&&\\0&\ddots &1&&&&&\\0&\ddots &0&1&&&&\vdots \\&&0&l_{i+1,i}&1&&&\vdots \\\vdots &&0&l_{i+2,i}&0&\ddots &&\\&&\vdots &\vdots &\vdots &\ddots &1&\\0&\dots &0&l_{n,i}&0&\dots &0&1\\\end{bmatrix}}}$

${\displaystyle \mathbf {L} _{i}^{-1}={\begin{bmatrix}1&&&\dots &&&&0\\0&\ddots &&&&&&\\0&\ddots &1&&&&&\\0&\ddots &0&1&&&&\vdots \\&&0&-l_{i+1,i}&1&&&\vdots \\\vdots &&0&-l_{i+2,i}&0&\ddots &&\\&&\vdots &\vdots &\vdots &\ddots &1&\\0&\dots &0&-l_{n,i}&0&\dots &0&1\\\end{bmatrix}}}$

## 向前与向后替换

### 向后替换

${\displaystyle {\begin{matrix}l_{1,1}x_{1}&&&&&=&b_{1}\\l_{2,1}x_{1}&+&l_{2,2}x_{2}&&&=&b_{2}\\\vdots &&\vdots &\ddots &&&\vdots \\l_{m,1}x_{1}&+&l_{m,2}x_{2}&+\dotsb +&l_{m,m}x_{m}&=&b_{m}\\\end{matrix}}}$

${\displaystyle x_{1}={\frac {b_{1}}{l_{1,1}}},}$
${\displaystyle x_{2}={\frac {b_{2}-l_{2,1}x_{1}}{l_{2,2}}},}$
${\displaystyle \vdots }$
${\displaystyle x_{m}={\frac {b_{m}-\sum _{i=1}^{m-1}l_{m,i}x_{i}}{l_{m,m}}}}$

## 注释与参考

1. Sheldon Axler. Linear Algebra Done Right. Springer-Verlag. 1996: 86–87, 169. ISBN 0-387-98258-2 （英语）.
2. ^ I. N. Herstein. Topics in Algebra 2. John Wiley and Sons. 1975: 285–290. ISBN 0-471-01090-1 （英语）.
3. ^ 这里要假设LU都可逆（对角线元素不为零），否则方程一般无解。
• 许以超. 线性代数与矩阵论 2. 高等教育出版社 （中文（中国大陆）‎）.