Woodbury matrix identity
woodbury matrix identity
2014/6/20
【转载请注明出处】http://www.cnblogs.com/mashiqi
http://en.wikipedia.org/wiki/Woodbury_matrix_identity
Today I'm going to write down a proof of this Woodbury matrix identity, which is very important in some practical situation. For instance, the 40th equation of this paper" bayesian compressive sensing using Laplace priors" applied this identity. Now let me give the details of it.
The Woodbury matrix identity is:
${(A + UCV)^{ - 1}} = {A^{ - 1}} - {A^{ - 1}}U{({C^{ - 1}} + V{A^{ - 1}}U)^{ - 1}}V{A^{ - 1}}$
where
,
,
and
are both assumed reversible.
Proof:
We denote
with
, namely
.So:
\[M{A^{ - 1}} = I + UCV{A^{ - 1}}\]
By multiply U with both side we get:
\[\begin{array}{l}
M{A^{ - 1}}U = U + UCV{A^{ - 1}}U = U(I + CV{A^{ - 1}}U)\\
{\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} = UC({C^{ - 1}} + V{A^{ - 1}}U)
\end{array}\]

is reversible,
we get:

But how could we deal with this nasty
term? We should notice that this term, which may not square, is coming from
itself, which is right a square and reversible matrix. So, from formula , we make up a pleasant
with is nasty
:
\[\begin{array}{l}
M{A^{ - 1}}U{({C^{ - 1}} + V{A^{ - 1}}U)^{ - 1}}V + A = UCV + A = M\\
\Rightarrow M = M{A^{ - 1}}U{({C^{ - 1}} + V{A^{ - 1}}U)^{ - 1}}V + A\\
\Rightarrow I - {A^{ - 1}}U{({C^{ - 1}} + V{A^{ - 1}}U)^{ - 1}}V = {M^{ - 1}}A
\end{array}\]
And finally due to the reversibility of
, we get the Woodbury matrix identity:
\[{M^{ - 1}} = {(A + VCU)^{ - 1}} = {A^{ - 1}} - {A^{ - 1}}U{({C^{ - 1}} + V{A^{ - 1}}U)^{ - 1}}V{A^{ - 1}}\]
Done.
We should notice that if
and
are identity matrix, then Woodbury matrix identity can be reduced to this form:
\[{(A + C)^{ - 1}} = {A^{ - 1}} - {A^{ - 1}}{({C^{ - 1}} + {A^{ - 1}})^{ - 1}}{A^{ - 1}}\]
,which is equivalent to:
\[{(A + C)^{ - 1}} = {C^{ - 1}}{({C^{ - 1}} + {A^{ - 1}})^{ - 1}}{A^{ - 1}}\]
This is because:
\[\begin{array}{l}
{(A + C)^{ - 1}} = {A^{ - 1}} - ( - {C^{ - 1}} + {C^{ - 1}} + {A^{ - 1}}){({C^{ - 1}} + {A^{ - 1}})^{ - 1}}{A^{ - 1}}\\
{\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} = {A^{ - 1}} + {C^{ - 1}}{({C^{ - 1}} + {A^{ - 1}})^{ - 1}}{A^{ - 1}} - ({C^{ - 1}} + {A^{ - 1}}){({C^{ - 1}} + {A^{ - 1}})^{ - 1}}{A^{ - 1}}\\
{\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} = {A^{ - 1}} + {C^{ - 1}}{({C^{ - 1}} + {A^{ - 1}})^{ - 1}}{A^{ - 1}} - {A^{ - 1}}\\
{\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} = {C^{ - 1}}{({C^{ - 1}} + {A^{ - 1}})^{ - 1}}{A^{ - 1}}
\end{array}\]
Woodbury matrix identity的更多相关文章
- 伍德伯里矩阵恒等式(Woodbury matrix identity)
宜言饮酒,与子偕老.琴瑟在御,莫不静好. 更多精彩内容请关注微信公众号 "优化与算法" 在数学(特别是线性代数)中,Woodbury矩阵恒等式是以Max A.Woodbury命名的 ...
- 最大比率传输(Maximum Ratio Transmission, MRT)原理分析
转载请注明出处. 最大比率发射(Maximum Ratio Transmission, MRT)是文献中经常看见的一个词,今天就在这里做一下笔记. 参考文献为:T. K. Y. Lo, "M ...
- 新基建新机遇!100页PPT
"新基建"是与传统的"铁公基"相对应,结合新一轮科技革命和产业变革特征,面向国家战略需求,为经济社会的创新.协调.绿色.开放.共享发展提供底层支撑的具有乘数效应 ...
- MIMO OFDM 常用信号检测算法
MIMO OFDM 系统检测算法 1. 前言 MIMO的空分复用技术可以使得系统在系统带宽和发射带宽不变的情况下容易地获得空间分集增益和信道的容量增益.OFDM技术采用多个正交的子载波并行传输数据,使 ...
- OSG中的HUD
OSG中的HUD 所谓HUD节点,说白了就是无论三维场景中的内容怎么改变,它都能在屏幕上固定位置显示的节点. 实现要点: 关闭光照,不受场景光照影响,所有内容以同一亮度显示 关闭深度测试 调整渲染顺序 ...
- osg实例介绍
osg实例介绍 转自:http://blog.csdn.net/yungis/article/list/1 [原]osgmotionblur例子 该例子演示了运动模糊的效果.一下内容是转自网上的:原理 ...
- OSG动画学习
OSG动画学习 转自:http://bbs.osgchina.org/forum.php?mod=viewthread&tid=3899&_dsign=2587a6a9 学习动画,看了 ...
- 重新想象 Windows 8 Store Apps (50) - 输入: 边缘手势, 手势操作, 手势识别
[源码下载] 重新想象 Windows 8 Store Apps (50) - 输入: 边缘手势, 手势操作, 手势识别 作者:webabcd 介绍重新想象 Windows 8 Store Apps ...
- osg 中鼠标拾取线段的端点和中点
//NeartestPointNodeVisitor.h #pragma once #include <osg\Matrix> #include <vector> #inclu ...
随机推荐
- 【leetcode❤python】 396. Rotate Function
#-*- coding: UTF-8 -*- #超时# lenA=len(A)# maxSum=[]# count=0# while count ...
- jQuery验证元素是否为空的两种常用方法
这篇文章主要介绍了jQuery验证元素是否为空的两种常用方法,实例分析了两种常用的判断为空技巧,非常具有实用价值,需要的朋友可以参考下 本文实例讲述了jQuery验证元素是否为空的两种常用方法.分享给 ...
- C语言运算符和优先级
关于C语言运算符和优先级,经整理众多博客资料汇入自己的实战,如下: a.算术运算 C语言一共有34种运算符,包括常见的加减乘除运算. 1) 加法:+ 还可以表 ...
- Shell 语法之函数
函数是被赋予名称的脚本代码块,可以在代码的任意位置重用.每当需要在脚本中使用这样的代码块时,只需引用该代码块被赋予的函数名称. 创建函数 格式 function name { commands } n ...
- spring aop搭建redis缓存
SpringAOP与Redis搭建缓存 近期项目查询数据库太慢,持久层也没有开启二级缓存,现希望采用Redis作为缓存.为了不改写原来代码,在此采用AOP+Redis实现. 目前由于项目需要,只需要做 ...
- [Selenium] 拖拽一个 Component 到 Workspace
先使Component可见,获取Component位置信息,获取Workspace位置信息,点击Component并拖拽到Workspace,最后释放.(调试时dragAndDropOffset()方 ...
- ASP.NET MVC Html.Partial/Html.RenderPartial/Html.Action/Html.RenderAction区别
1. @Html.Raw() 方法输出带有html标签的字符串: <div style="margin:10px 0px 0px;border:1px;border-color:red ...
- mysql 处理查询请求过程
需要搞清楚查询为什么会慢,就要搞清楚mysql处理查询请求的过程: 1.客户端发送SQL请求给服务器 2.服务器检查是否可以在查询缓存中命中该SQL 查询缓存对SQL性能的影响. 1.需要对缓存加 ...
- Microsoft Enterprise Library 5.0 缓存配置
在使用企业库的缓存时遇到一个问题. 创建 cachingConfiguration 的配置节处理程序时出错: 未能加载文件或程序集“Microsoft.Practices.EnterpriseLibr ...
- WCF初探-15:WCF操作协定
前言: 在前面的文章中,我们定义服务协定时,在它的操作方法上都会加上OperationContract特性,此特性属于OperationContractAttribute 类,将OperationCo ...