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<record version="11" id="1287">
 <title>Cholesky decomposition</title>
 <name>CholeskyDecomposition</name>
 <created>2002-01-05 00:25:42</created>
 <modified>2006-06-04 15:46:01</modified>
 <type>Definition</type>
 <creator id="12050" name="gufotta"/>
 <author id="12050" name="gufotta"/>
 <author id="2" name="akrowne"/>
 <classification>
	<category scheme="msc" code="15-00"/>
	<category scheme="msc" code="65-00"/>
	<category scheme="msc" code="62J05"/>
 </classification>
 <defines>
	<concept>Cholesky triangle</concept>
 </defines>
 <synonyms>
	<synonym concept="Cholesky decomposition" alias="Cholesky factorization"/>
	<synonym concept="Cholesky decomposition" alias="matrix square root"/>
 </synonyms>
 <related>
	<object name="SquareRootOfPositiveDefiniteMatrix"/>
 </related>
 <keywords>
	<term>matrix factorizations</term>
 </keywords>
 <preamble>\usepackage{amssymb}
\usepackage{amsmath}
\usepackage{amsfonts}
%\usepackage{graphicx}
%\usepackage{xypic}</preamble>
 <content>\section{Cholesky Decomposition}

A symmetric and positive definite matrix can be efficiently decomposed into a lower and upper triangular matrix. For a matrix of any type, this is achieved by the LU decomposition which factorizes $A = LU$. If $A$ satisfies the above criteria, one can decompose more efficiently into $A=LL^T$ where $L$ is a lower triangular matrix with positive diagonal elements. $L$ is called the \emph{Cholesky triangle}.

To solve $Ax = b$, one solves first $Ly = b$ for $y$, and then $L^Tx=y$ for $x$.

A variant of the Cholesky decomposition is the form $A=R^TR$ , where $R$ is upper triangular.

Cholesky decomposition is often used to solve the normal equations in linear least squares problems; they give $A^TAx=A^Tb$ , in which $A^TA$ is symmetric and positive definite.

To derive $A=LL^T$, we simply equate coefficients on both sides of the equation:

$$ 
\begin{bmatrix}
 a_{11} &amp; a_{12} &amp; \cdots &amp; a_{1n} \\
 a_{21} &amp; a_{22} &amp; \cdots &amp; a_{2n} \\
 a_{31} &amp; a_{32} &amp; \cdots &amp; a_{3n} \\
 \vdots &amp; \vdots &amp; \ddots &amp; \vdots \\
 a_{n1} &amp; a_{n2} &amp; \cdots &amp; a_{nn} 
\end{bmatrix}
=
\begin{bmatrix}
 l_{11} &amp; 0      &amp; \cdots &amp; 0 \\
 l_{21} &amp; l_{22} &amp; \cdots &amp; 0 \\
 \vdots &amp; \vdots &amp; \ddots &amp; \vdots \\
 l_{n1} &amp; l_{n2} &amp; \cdots &amp; l_{nn}
\end{bmatrix}
\begin{bmatrix}
 l_{11} &amp; l_{21} &amp; \cdots &amp; l_{n1} \\
 0      &amp; l_{22} &amp; \cdots &amp; l_{n2} \\
 \vdots &amp; \vdots &amp; \ddots &amp; \vdots \\
 0      &amp; 0      &amp; \cdots &amp; l_{nn}
\end{bmatrix}
$$ 

Solving for the unknowns (the nonzero $l_{ji}$s), for $i=1,\cdots ,n$ and $j=i+1,\ldots ,n$, we get:

\begin{eqnarray*}
 l_{ii} &amp; = &amp; \sqrt{\left( a_{ii} - \sum_{k=1}^{i-1} l_{ik}^2 \right)} \\
 l_{ji} &amp; = &amp; \left(a_{ji} - \sum_{k=1}^{i-1} l_{jk} l_{ik} \right) / l_{ii}
\end{eqnarray*}

Because $A$ is symmetric and positive definite, the expression under the square root is always positive, and all $l_{ij}$ are real. 

\begin{thebibliography}{3}

\bibitem{DAB} Originally from The Data Analysis Briefbook
(\PMlinkexternal{http://rkb.home.cern.ch/rkb/titleA.html}{http://rkb.home.cern.ch/rkb/titleA.html})

\end{thebibliography}</content>
</record>
