“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
-
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 ...
-
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 ...
-
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 ...
-
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 ...
-
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 ...
-
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 ...
-
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 ...
-
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 ...
-
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
-
强大的Resharp插件
使用VS有段时间了,一直深深的折服于其强大的功能.之前一直听说有Resharp这个工具,小猪一直也没有太在意.直到今天…… 下载安装: http://www.jetbrains.com/resharp ...
-
arcgis中求多点到一条曲线的最短欧几里得距离
1.使用的工具:Arctoolbox----Analysis Tools----Proximity----Near工具. 2.注意:在求距离之前一定要先设置好坐标系统.
-
Redis客户端Java服务接口封装
最近在学习Redis并集成到Spring中去,发现Spring的RedisTemplate并不好用,还没有MongoTemplate好用. 而且发现Jedis和ShardedJedis的方法非常多,覆 ...
-
Servlet &; JSP - Cookie
关于 Cookie 的内容,参考 HTTP - Cookie 机制 获取来自客户端的 cookie request.getCookies 方法可以获取来自 HTTP 请求的 cookie,返回的是 j ...
-
NET异步调用Webserver
之前,有个同事跑来问我一堆的什么多线程异步进行调用Sap的服务再突然把进程关闭,还说要设置一个循环判断调用的结果,搞得我听的一头雾水,但是我明显感觉到他的设计思路已经渐行渐远了...已经再偏远的山区中 ...
-
Math.random与java.util.Random的差别
今天在做一道习题时想到了Math.random()与Random类有什么区别,查阅了一些资料,感觉讲的不是太好. 首先两者的区别是一个是方法,一个是类. 其实前者的实现借助与后者.大家可以看一下Mat ...
-
linux(4) vi编辑/删除、复制、粘贴 /bash shell 环境变量设置/数据流重定向 | 的用法
一.vi文字处理器1.vi与vimvi:文字处理器vim:程序开发工具2.vi介绍三种模式:一般模式(vi刚进入的,不可编辑),编辑模式(按i后,左下方是insert)和命令行模式(按esc退出,:w ...
-
大数据 - Teradata学习体会
引言 随着计算机系统在处理能力.存储能力等方面,特别是计算机软件技术的不断提高,使得信息处理技术得到飞速发展. 数据处理主要分为两大类:联机事物处理OLTP.联机分析处理OLAP.OLTP也就是传统的 ...
-
python模拟艺龙网登录带验证码输入
1.使用urllib与urllib2包 2.使用cookielib自动管理cookie 3.360浏览器F12抓信息 登录请求地址和验证码地址都拿到了如图 # -*- coding: utf-8 -* ...
-
用开源项目PhotoView实现图片的双指缩放和双击放大缩小
项目地址:https://github.com/chrisbanes/PhotoView 用开源项目有个好处,一是实现简单,二是bug少.那么我们就来说下这个项目能够实现的效果: 1.单个图片的双指缩 ...