\documentclass[twoside]{article}

\usepackage{palatino}
\usepackage{camnum}

\textheight=22.5cm
\textwidth=15cm
\topmargin=-1cm
\oddsidemargin=0.5cm
\evensidemargin=0.5cm
\parindent=0pt
\parskip=6pt

\begin{document}
\begin{center}
  {\large \textbf{Mathematical Tripos Part IB: Easter 2001}}

  {\Large \textbf{Numerical Analysis -- Detailed Example
      6a\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 QR factorization: Householder reflections}}
\end{center}

Let
\begin{displaymath}
   A
 = \left[\begin{array}{rrr}
      2 &  4 & 5\\
      1 & -1 & 1\\
      2 &  1 &-1
    \end{array}\right].
\end{displaymath}
Then
$$
\begin{array}{llr}
   \MM{u}_1
 = \left[\begin{array}{r}
      -1\\1\\2
    \end{array}\right],\qquad
%
&   \MM{u}_1\MM{u}_1^{\CC{T}}
 = \left[\begin{array}{rrr}
       1 & -1 & -2 \\
      -1 &  1 &  2 \\
      -2 &  2 &  4
    \end{array}\right],\quad
%
&   \Omega_1
 = I-2\frac{\MM{u}_1\MM{u}_1^{\CC{T}}}{\|\MM{u}_1\|^2}
 = \frac13
   \left[\begin{array}{rrr}
       2 &  1 &  2 \\
       1 &  2 & -2 \\
       2 & -2 & -1
    \end{array}\right],\\[4ex]
%
&&   R_1 = \Omega_1 A
  = \left[\begin{array}{rrr}
       3 &  \phantom{-}3 &  \phantom{-}3 \\
       0 &  0 & 3 \\
       0 &  3 & 3
    \end{array}\right],\\[4ex]
%%
   \MM{u}_2
 = \left[\begin{array}{r}
      0\\-1\\1
    \end{array}\right],\quad
%
&   \MM{u}_2\MM{u}_2^{\CC{T}}
 = \left[\begin{array}{rrr}
       \phantom{-}0 &  0 &  0 \\
      0 &  1 & -1 \\
      0 & -1 &  1
    \end{array}\right],\quad
%
&   \Omega_2
 = I-2\frac{\MM{u}_2\MM{u}_2^{\CC{T}}}{\|\MM{u}_2\|^2}
 = 
   \left[\begin{array}{rrr}
       1 &   \phantom{-}0 &   \phantom{-}0 \\
       0 &  0 &  1 \\
       0 &  1 &  0
    \end{array}\right],\\[4ex]
%
&&   R  = \Omega_2 R_1
  = \left[\begin{array}{rrr}
       3 &   \phantom{-}3 &  \phantom{-}3 \\
       0 &  3 & 3 \\
       0 &  0 & 3
    \end{array}\right].
\end{array}
$$
Finally,
\begin{displaymath}
   Q = (\Omega_2\Omega_1)^{\CC{T}}
 = \left(\left[\begin{array}{rrr}
       1 &  0 &  0 \\
       0 &  0 &  1 \\
       0 &  1 &  0
    \end{array}\right]
   \frac13
   \left[\begin{array}{rrrr}
       2 &  1 &  2 \\
       1 &  2 & -2 \\
       2 & -2 & -1
    \end{array}\right]\right)^{\CC{T}}
 = \frac13
   \left[\begin{array}{rrr}
       2 &  1 &  2 \\
       1 & -2 & -1 \\
       2 &  2 & -2
    \end{array}\right]^{\CC{T}}
 = \frac13
   \left[\begin{array}{rrr}
       2 &  2 &  1 \\
       1 & -2 &  2 \\
       2 & -1 & -2
    \end{array}\right]
\end{displaymath}

\end{document}
%%% Local Variables: 
%%% mode: latex
%%% TeX-master: t
%%% End: 
