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  {\large \textbf{Mathematical Tripos Part IB: Easter 20001}}

  {\Large \textbf{Numerical Analysis -- Detailed Example
      3\footnote{Corrections and suggestions to these notes should
        be emailed to \texttt{a.shadrin@damtp.cam.ac.uk}. All handouts
        are available on the WWW at the URL 
        \texttt{http://www.damtp.cam.ac.uk/user/na/PartIB/Handouts.html}.}}}

  \vspace{8pt}
  \textbf{\textit{\LARGE Cholesky factorization}}
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Let
\begin{displaymath}
  A=\left[
    \begin{array}{ccccc}
      1 & 2 & 0 & 0 & 0\\
      2 & 5 & 2 & 0 & 0\\
      0 & 3 & 10 & 4 & 0\\
      0 & 0 & 4 & 17 & 5\\
      0 & 0 & 0 & 5 & 26
    \end{array}\right],
\end{displaymath}
hence $A$ is both symmetric (actually, as we'll see, positive
definite) and banded. Using the matrix itself to store columns of $L$
as we form them, we have
\begin{Eqnarray*}
  &&\MM{\ell}_1=\left[
    \begin{array}{c}
      1\\2\\0\\0\\0
    \end{array}\right],\qquad A=\left[
    \begin{array}{c|cccc}
      1 & 2 & 0 & 0 & 0\\\hline
      2 & 1 & 3 & 0 & 0\\
      0 & 3 & 10 & 4 & 0\\
      0 & 0 & 4 & 17 & 5\\
      0 & 0 & 0 & 5 & 26
    \end{array}\right],\\
  &&\MM{\ell}_2=\left[
    \begin{array}{c}
      0\\1\\3\\0\\0
    \end{array}\right],\qquad A=\left[
    \begin{array}{cc|ccc}
      1 & 2 & 0 & 0 & 0\\
      2 & 1 & 3 & 0 & 0\\\hline
      0 & 3 & 1 & 4 & 0\\
      0 & 0 & 4 & 17 & 5\\
      0 & 0 & 0 & 5 & 26
    \end{array}\right],\\
  &&\MM{\ell}_3=\left[
    \begin{array}{c}
      0\\0\\1\\4\\0
    \end{array}\right],\qquad A=\left[
    \begin{array}{ccc|cc}
      1 & 2 & 0 & 0 & 0\\
      2 & 1 & 3 & 0 & 0\\
      0 & 3 & 1 & 4 & 0\\\hline
      0 & 0 & 4 & 1 & 5\\
      0 & 0 & 0 & 5 & 26
    \end{array}\right],\\
  &&\MM{\ell}_4=\left[
    \begin{array}{c}
      0\\0\\0\\1\\5
    \end{array}\right],\qquad A=\left[
    \begin{array}{cccc|c}
      1 & 2 & 0 & 0 & 0\\
      2 & 1 & 3 & 0 & 0\\
      0 & 3 & 1 & 4 & 0\\
      0 & 0 & 4 & 1 & 5\\
      0 & 0 & 0 & 5 & 1
    \end{array}\right],\\
  &&\MM{\ell}_5=\left[
    \begin{array}{c}
      0\\0\\0\\0\\1
    \end{array}\right]\qquad\Rightarrow\qquad A=LL^{\CC{T}},\qquad L=\left[
    \begin{array}{ccccc}
      1 & 0 & 0 & 0 & 0\\
      2 & 1 & 0 & 0 & 0\\
      0 & 3 & 1 & 0 & 0\\
      0 & 0 & 4 & 1 & 0\\
      0 & 0 & 0 & 5 & 1
    \end{array}\right].
\end{Eqnarray*}


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