“矩阵代数初步”(Introduction to MATRIX ALGEBRA)课程由Prof. A.K.Kaw(University of South Florida)设计并讲授。
PDF格式学习笔记下载(Academia.edu)
第7章课程讲义下载(PDF)

Summary

  • For a nonsingular matrix $[A]$ on which one can always write it as $$[A] = [L][U]$$ where $[L]$ is a lower triangular matrix, $[U]$ is a upper triangular matrix.
  • Note that not all matrices have LU decomposition, such as $\begin{bmatrix}0& 2\\ 2& 0\end{bmatrix}$. $$\begin{bmatrix}0& 2\\ 2& 0\end{bmatrix}=\begin{bmatrix}1& 0\\ a& 1\end{bmatrix} \begin{bmatrix}b& c\\ 0& d\end{bmatrix} \Rightarrow \begin{cases} b=0\\ ab=2\end{cases}$$ which is contradiction.
  • If one is solving a set of equations $$[A][X]=[B]$$ then $$LUX=B$$ $$\Rightarrow L^{-1}LUX=L^{-1}B$$ $$\Rightarrow UX=L^{-1}B=Y$$ then we have $$\begin{cases}LY=B\\ UX=Y\end{cases}$$ So we can solve the first equation for $[Y]$by using forward substitution and then use the second equation to calculate the solution vector $[X]$ by back substitution.
  • For instance, solve the following set of equations: $$\begin{bmatrix}1& 2& 3\\ 2& 1& -4\\ 1& 5& 2\end{bmatrix}\cdot \begin{bmatrix} x\\ y\\ z \end{bmatrix} = \begin{bmatrix} 14\\ -8\\ 17\end{bmatrix}$$ Applying LU decomposition on the coefficient matrix,
    • Firstly write down an identity matrix (the same size as the coefficient matrix) on the left and the coefficient matrix on the right. $$L\leftarrow\begin{bmatrix}1& 0& 0\\ 0& 1& 0\\ 0& 0& 1 \end{bmatrix} \begin{bmatrix}1& 2& 3\\ 2& 1& -4\\ 1& 5& 2\end{bmatrix}\rightarrow U$$
    • Then applying elementary row operation on the right while simultaneously updating successive columns of the matrix on the left. For example, if we are doing $R_1 + m R_2$ on the right then we will do $C_2-mC_1$ on the left. That is, we will keep the equivalent of the product. $$\begin{bmatrix}1& 0& 0\\ 0& 1& 0\\ 0& 0& 1 \end{bmatrix} \begin{bmatrix}1& 2& 3\\ 2& 1& -4\\ 1& 5& 2\end{bmatrix}$$ $$\Rightarrow\begin{cases}R_2-2R_1 \\ C_1+2C_2\end{cases} \begin{bmatrix}1& 0& 0\\ 2& 1& 0\\ 0& 0& 1 \end{bmatrix} \begin{bmatrix}1& 2& 3\\ 0& -3& -10\\ 1& 5& 2\end{bmatrix}$$ $$\Rightarrow\begin{cases}R_3-R_1 \\ C_1+C_3\end{cases} \begin{bmatrix}1& 0& 0\\ 2& 1& 0\\ 1& 0& 1 \end{bmatrix} \begin{bmatrix}1& 2& 3\\ 0& -3& -10\\ 0& 3& -1\end{bmatrix}$$ $$\Rightarrow\begin{cases}R_3+R_2 \\ C_2-C_3\end{cases} \begin{bmatrix}1& 0& 0\\ 2& 1& 0\\ 1& -1& 1 \end{bmatrix} \begin{bmatrix}1& 2& 3\\ 0& -3& -10\\ 0& 0& -11\end{bmatrix}$$ Thus far, the right matrix is an upper triangular matrix (i.e. $U$) and the left one is a lower triangular matrix (i.e. $L$).
    • Solving $[L][Y]=[B]$, that is $$\begin{bmatrix}1& 0& 0\\ 2& 1& 0\\ 1& -1& 1 \end{bmatrix}\cdot Y=\begin{bmatrix} 14\\ -8\\ 17\end{bmatrix}\Rightarrow Y=\begin{bmatrix}14\\ -36\\ -33\end{bmatrix}$$
    • Solving $[U][X]=[Y]$, that is $$\begin{bmatrix}1& 2& 3\\ 0& -3& -10\\ 0& 0& -11\end{bmatrix}\cdot \begin{bmatrix} x\\ y\\ z \end{bmatrix} = \begin{bmatrix}14\\ -36\\ -33\end{bmatrix}$$ $$ \Rightarrow\begin{cases}x=1\\ y=2 \\ z=3\end{cases}$$

Selected Problems

1. Find the $[L]$ and $[U]$ matrices of the following matrix $$\begin{bmatrix}25& 5& 4\\ 75& 7& 16\\ 12.5& 12& 22 \end{bmatrix}$$

Solution:
$$\begin{bmatrix}1& 0& 0\\ 0& 1& 0\\ 0& 0& 1 \end{bmatrix}\begin{bmatrix}25& 5& 4\\ 75& 7& 16\\ 12.5& 12& 22 \end{bmatrix}$$ $$\Rightarrow \begin{cases}R_2-3R_1\\ R_3-{1\over2}R_1\\ C_1+3C_2\\ C_1+{1\over2}C_3\end{cases} \begin{bmatrix}1& 0& 0\\ 3& 1& 0\\ {1\over2}& 0& 1 \end{bmatrix} \begin{bmatrix}25& 5& 4\\ 0& -8& 4\\ 0& 9.5& 20 \end{bmatrix}$$ $$\Rightarrow \begin{cases}R_3+{19\over16}R_2\\C_2-{19\over16}C_3\end{cases} \begin{bmatrix}1& 0& 0\\ 3& 1& 0\\ {1\over2}& -{19\over16}& 1 \end{bmatrix} \begin{bmatrix}25& 5& 4\\ 0& -8& 4\\ 0& 0& {99\over4} \end{bmatrix}$$ That is, $$L= \begin{bmatrix}1& 0& 0\\ 3& 1& 0\\ {1\over2}& -{19\over16}& 1 \end{bmatrix},\ U = \begin{bmatrix}25& 5& 4\\ 0& -8& 4\\ 0& 0& {99\over4} \end{bmatrix}.$$

2. Using LU decomposition to solve: $$\begin{cases} 4x_1 + x_2 - x_3 = -2\\ 5x_1+x_2+2x_3=4\\ 6x_1+x_2+x_3=6 \end{cases}$$

Solution:
$$\begin{bmatrix}1& 0& 0\\ 0& 1& 0\\ 0& 0& 1 \end{bmatrix} \begin{bmatrix}4& 1& -1\\ 5& 1& 2\\ 6& 1& 1\end{bmatrix}$$ $$\Rightarrow \begin{cases}R_2-{5\over4}R_1\\ R_3-{3\over2}R_1\\ C_1+{5\over4}C_2\\ C_1+{3\over2}C_3\end{cases} \begin{bmatrix}1& 0& 0\\ {5\over4}& 1& 0\\ {3\over2}& 0& 1 \end{bmatrix} \begin{bmatrix}4& 1& -1\\ 0& -{1\over4}& {13\over4}\\ 0& -{1\over2}& {5\over2}\end{bmatrix}$$ $$\Rightarrow \begin{cases}R_3-2R_2\\ C_2+2C_3\end{cases} \begin{bmatrix}1& 0& 0\\ {5\over4}& 1& 0\\ {3\over2}& 2& 1 \end{bmatrix} \begin{bmatrix}4& 1& -1\\ 0& -{1\over4}& {13\over4}\\ 0&0& -4\end{bmatrix}$$ That is, $$L = \begin{bmatrix}1& 0& 0\\ {5\over4}& 1& 0\\ {3\over2}& 2& 1 \end{bmatrix},\ U= \begin{bmatrix}4& 1& -1\\ 0& -{1\over4}& {13\over4}\\ 0&0& -4\end{bmatrix}.$$ Then we solve $[L][Y]=[B]$, $$\begin{bmatrix}1& 0& 0\\ {5\over4}& 1& 0\\ {3\over2}& 2& 1 \end{bmatrix}\cdot Y=\begin{bmatrix}-2\\4\\6\end{bmatrix} \Rightarrow Y=\begin{bmatrix}-2\\{13\over2}\\ -4\end{bmatrix}$$ Finally, we solve $[U][X]=[Y]$, $$\begin{bmatrix}4& 1& -1\\ 0& -{1\over4}& {13\over4}\\ 0&0& -4\end{bmatrix}\cdot X= \begin{bmatrix}-2\\{13\over2}\\ -4\end{bmatrix}\Rightarrow X=\begin{bmatrix}3\\-13\\1\end{bmatrix}$$ Thus the solution is $$\begin{cases}x_1 = 3\\ x_2 = -13\\ x_3 = 1\end{cases}$$

3. Find the inverse of $$[A]=\begin{bmatrix}3& 4& 1\\ 2& -7& -1\\ 8& 1& 5\end{bmatrix}$$

Solution:

To find the inverse of a matrix, actually it is to solve a set of equations: $$\begin{cases}AX_1=[1, 0, 0]^{T}\\ AX_2 = [0, 1, 0]^{T}\\ AX_3 = [0, 0, 1]^{T} \end{cases}$$ Firstly, we will find the $[L]$ and $[U]$. $$\begin{bmatrix}1& 0& 0\\ 0& 1& 0\\ 0& 0& 1 \end{bmatrix} \begin{bmatrix}3& 4& 1\\ 2& -7& -1\\ 8& 1& 5\end{bmatrix}$$ $$\Rightarrow \begin{cases}R_2-{2\over3}R_1\\ R_3-{8\over3}R_1\\ C_1+{2\over3}C_2\\ C_1+{8\over3}C_3\end{cases} \begin{bmatrix}1& 0& 0\\ {2\over3}& 1& 0\\ {8\over3}& 0& 1 \end{bmatrix} \begin{bmatrix}3& 4& 1\\ 0& -{29\over3}& -{5\over3}\\ 0& -{29\over3}& {7\over3}\end{bmatrix}$$ $$\Rightarrow \begin{cases}R_3-R_2\\ C_2+C_3\end{cases} \begin{bmatrix}1& 0& 0\\ {2\over3}& 1& 0\\ {8\over3}& 1& 1 \end{bmatrix} \begin{bmatrix}3& 4& 1\\ 0& -{29\over3}& -{5\over3}\\ 0&0& 4\end{bmatrix}$$ That is, $$L = \begin{bmatrix}1& 0& 0\\ {2\over3}& 1& 0\\ {8\over3}& 1& 1 \end{bmatrix},\ U= \begin{bmatrix}3& 4& 1\\ 0& -{29\over3}& -{5\over3}\\ 0&0& 4\end{bmatrix}.$$ Then we solve $[L][Y]=[I]$, note that there are three columns of $[Y]$: $$LY_1 = \begin{bmatrix}1& 0& 0\\ {2\over3}& 1& 0\\ {8\over3}& 1& 1 \end{bmatrix} \cdot Y_1 = \begin{bmatrix}1\\0\\0\end{bmatrix} \Rightarrow Y_1=\left[1, -{2\over3}, -2\right]^{T}$$ $$LY_2 = \begin{bmatrix}1& 0& 0\\ {2\over3}& 1& 0\\ {8\over3}& 1& 1 \end{bmatrix} \cdot Y_2 = \begin{bmatrix}0\\1\\0\end{bmatrix} \Rightarrow Y_2=\left[0, 1, -1\right]^{T}$$ $$LY_3 = \begin{bmatrix}1& 0& 0\\ {2\over3}& 1& 0\\ {8\over3}& 1& 1 \end{bmatrix} \cdot Y_3 = \begin{bmatrix}0\\0\\1\end{bmatrix} \Rightarrow Y_3=\left[0, 0, 1\right]^{T}$$ Finally we can solve $[X]$ by $[U][X]=[Y]$: $$UX_1=Y_1\Rightarrow \begin{bmatrix}3& 4& 1\\ 0& -{29\over3}& -{5\over3}\\ 0&0& 4\end{bmatrix} \cdot X_1 = \begin{bmatrix}1\\ -{2\over3}\\ -2\end{bmatrix}\Rightarrow X_1= \left[{17\over58}, {9\over58}, -{1\over2}\right]^{T}$$ $$UX_2=Y_2\Rightarrow \begin{bmatrix}3& 4& 1\\ 0& -{29\over3}& -{5\over3}\\ 0&0& 4\end{bmatrix} \cdot X_2 = \begin{bmatrix}0\\ 1\\ -1\end{bmatrix}\Rightarrow X_2= \left[{19\over116}, -{7\over116}, -{1\over4}\right]^{T}$$ $$UX_3=Y_3\Rightarrow \begin{bmatrix}3& 4& 1\\ 0& -{29\over3}& -{5\over3}\\ 0&0& 4\end{bmatrix} \cdot X_3 = \begin{bmatrix}0\\ 0\\ 1\end{bmatrix}\Rightarrow X_3= \left[-{3\over116}, -{5\over116}, {1\over4}\right]^{T}$$ Thus the inverse of the original matrix is $$[A]^{-1} = \begin{bmatrix}{17\over58} & {19\over116} & -{3\over116}\\ {9\over58} & -{7\over116} & -{5\over116}\\ -{1\over2} & -{1\over4} & {1\over4}\end{bmatrix}$$

A.Kaw矩阵代数初步学习笔记 7. LU Decomposition的更多相关文章

  1. A.Kaw矩阵代数初步学习笔记 10. Eigenvalues and Eigenvectors

    “矩阵代数初步”(Introduction to MATRIX ALGEBRA)课程由Prof. A.K.Kaw(University of South Florida)设计并讲授. PDF格式学习笔 ...

  2. A.Kaw矩阵代数初步学习笔记 9. Adequacy of Solutions

    “矩阵代数初步”(Introduction to MATRIX ALGEBRA)课程由Prof. A.K.Kaw(University of South Florida)设计并讲授. PDF格式学习笔 ...

  3. A.Kaw矩阵代数初步学习笔记 8. Gauss-Seidel Method

    “矩阵代数初步”(Introduction to MATRIX ALGEBRA)课程由Prof. A.K.Kaw(University of South Florida)设计并讲授. PDF格式学习笔 ...

  4. A.Kaw矩阵代数初步学习笔记 6. Gaussian Elimination

    “矩阵代数初步”(Introduction to MATRIX ALGEBRA)课程由Prof. A.K.Kaw(University of South Florida)设计并讲授. PDF格式学习笔 ...

  5. A.Kaw矩阵代数初步学习笔记 5. System of Equations

    “矩阵代数初步”(Introduction to MATRIX ALGEBRA)课程由Prof. A.K.Kaw(University of South Florida)设计并讲授. PDF格式学习笔 ...

  6. A.Kaw矩阵代数初步学习笔记 4. Unary Matrix Operations

    “矩阵代数初步”(Introduction to MATRIX ALGEBRA)课程由Prof. A.K.Kaw(University of South Florida)设计并讲授. PDF格式学习笔 ...

  7. A.Kaw矩阵代数初步学习笔记 3. Binary Matrix Operations

    “矩阵代数初步”(Introduction to MATRIX ALGEBRA)课程由Prof. A.K.Kaw(University of South Florida)设计并讲授. PDF格式学习笔 ...

  8. A.Kaw矩阵代数初步学习笔记 2. Vectors

    “矩阵代数初步”(Introduction to MATRIX ALGEBRA)课程由Prof. A.K.Kaw(University of South Florida)设计并讲授. PDF格式学习笔 ...

  9. A.Kaw矩阵代数初步学习笔记 1. Introduction

    “矩阵代数初步”(Introduction to MATRIX ALGEBRA)课程由Prof. A.K.Kaw(University of South Florida)设计并讲授. PDF格式学习笔 ...

随机推荐

  1. JavaScript:关于事件处理程序何时可以直接访问元素的属性

    指定在元素的的事件处理程序中指定 <input type="button" value="click me" onclick="alert(th ...

  2. typedef 和define的区别

    总结一下typedef和#define的区别 1.概念 #define 它在编译预处理时进行简单的替换,不作正确性检查.它是预处理指令. typedef 它在自己的作用域内给一个已经存在的类型一个别名 ...

  3. 移动端页面(css)调试之“weinre大法”

    移动端页面调试一般分两步.第一步我们需要把本地(pc端)写的页面效果展现在移动端,一个很方便的办法是用 fiddler 作为代理,具体可以参考 如何用 fiddler 代理调试本地手机页面,这样我们就 ...

  4. unity3d CarWaypoints插件

    编写初衷: 1.网上没有现成的好用的waypoints插件 2.自己在做一个赛车游戏,如果没有这款插件的话在制作游戏的过程中会被累成狗 3.从来没有接触过插件方面的东西,所以想自己尝试一下 插件用途: ...

  5. Javascript 模块化开发上线解决方案

    最近又换部门了,好频繁地说...于是把这段时间搞的小工具们简单整理了一下,作了一个小的总结.这次用一个简单业务demo来向大家介绍一下Javascript模块化开发的方式和自动化合并压缩的一些自己的处 ...

  6. [BZOJ1193][HNOI2006]马步距离(贪心+dfs)

    题目:http://www.lydsy.com:808/JudgeOnline/problem.php?id=1193 分析: 首先小范围可以直接暴力.(其实只要用上题目中的表就行了) 如果范围比较大 ...

  7. 关于HTML+CSS设置图片居中的方法

    有的时候我们会遇到这样一个问题,就是当我们按下F12进行使用firebug调试的时候,我们发现图像没有居中,页面底下有横向的滑动条出现,图片没能够居中,默认状态下只是紧靠在页面最左侧,而我们对图像缩小 ...

  8. jQuery插件---exselect实现联动

    <!DOCTYPE HTML> <html> <head> <meta charset="utf-8"> <title> ...

  9. java/c# 判断点是否在多边形区域内

    java/c# 判断点是否在多边形区域内 年06月29日 ⁄ 综合 ⁄ 共 1547字 ⁄ 字号 小 中 大 ⁄ 评论关闭 最近帮别人解决了一个问题,如何判断一个坐标点,是否在多边形区域内(二维). ...

  10. 1117Mysql prepare预处理语句

    转自http://www.jb51.net/article/81378.htm 综述:一般用来拼凑SQL然后执行 MySQL 5.1对服务器一方的预制语句提供支持.如果您使用合适的客户端编程界面,则这 ...