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<H3><A NAME="SECTION03242300000000000000">
<B><I>QR</I></B> Factorization with Column Pivoting</A>
</H3>
<P>
To solve a linear least squares problem (<A HREF="node27.html#llsq">2.1</A>)<A NAME="2673"></A><A NAME="2674"></A>
when <B><I>A</I></B> is not of full rank, or the rank of <B><I>A</I></B> is in doubt, we can
perform either a <B><I>QR</I></B> factorization with column pivoting
<A NAME="2675"></A> or a singular value
decomposition (see subsection <A HREF="node53.html#subseccompsvd">2.4.6</A>).
<P>
The <B><I>QR</I></B> <B>factorization with column pivoting</B> is given by
<BR><P></P>
<DIV ALIGN="CENTER">
<!-- MATH
\begin{displaymath}
A = Q\left( \begin{array}{c}R\\0\end{array}\right)P^T, \quad m \ge n,
\end{displaymath}
-->
<IMG
WIDTH="212" HEIGHT="54" BORDER="0"
SRC="img111.png"
ALT="\begin{displaymath}
A = Q\left( \begin{array}{c}R\\ 0\end{array}\right)P^T, \quad m \ge n,
\end{displaymath}">
</DIV>
<BR CLEAR="ALL">
<P></P>
where <B><I>Q</I></B> and <B><I>R</I></B> are as before and <B><I>P</I></B> is a permutation matrix, chosen
(in general) so that
<BR><P></P>
<DIV ALIGN="CENTER">
<!-- MATH
\begin{displaymath}
|r_{11}| \ge |r_{22}| \ge \ldots \ge |r_{nn}|
\end{displaymath}
-->
<IMG
WIDTH="186" HEIGHT="31" BORDER="0"
SRC="img112.png"
ALT="\begin{displaymath}
\vert r_{11}\vert \ge \vert r_{22}\vert \ge \ldots \ge \vert r_{nn}\vert
\end{displaymath}">
</DIV>
<BR CLEAR="ALL">
<P></P>
and moreover, for each <B><I>k</I></B>,
<BR><P></P>
<DIV ALIGN="CENTER">
<!-- MATH
\begin{displaymath}
|r_{kk}| \ge \|R_{k:j,j}\|_2 \quad \mbox{for $j = k+1, \ldots, n$.}
\end{displaymath}
-->
<IMG
WIDTH="289" HEIGHT="32" BORDER="0"
SRC="img113.png"
ALT="\begin{displaymath}
\vert r_{kk}\vert \ge \Vert R_{k:j,j}\Vert _2 \quad \mbox{for $j = k+1, \ldots, n$.}
\end{displaymath}">
</DIV>
<BR CLEAR="ALL">
<P></P>
In exact arithmetic, if
<!-- MATH
$\mbox{rank}(A) = k$
-->
<IMG
WIDTH="98" HEIGHT="34" ALIGN="MIDDLE" BORDER="0"
SRC="img114.png"
ALT="$\mbox{rank}(A) = k$">,
then the whole of the submatrix
<B><I>R</I><SUB>22</SUB></B> in rows and columns <B><I>k</I>+1</B> to <B><I>n</I></B>
would be zero. In numerical computation, the aim must be to
determine an index <B><I>k</I></B>, such that the leading submatrix <B><I>R</I><SUB>11</SUB></B> in the first
<B><I>k</I></B> rows and columns is well-conditioned, and <B><I>R</I><SUB>22</SUB></B> is negligible:
<BR><P></P>
<DIV ALIGN="CENTER">
<!-- MATH
\begin{displaymath}
R = \left( \begin{array}{cc}R_{11} & R_{12} \\0 & R_{22} \end{array}\right)
\simeq \left( \begin{array}{cc}R_{11} & R_{12} \\0 & 0\end{array}\right) .
\end{displaymath}
-->
<IMG
WIDTH="291" HEIGHT="54" BORDER="0"
SRC="img115.png"
ALT="\begin{displaymath}
R = \left( \begin{array}{cc}R_{11} & R_{12} \\ 0 & R_{22} \e...
... \begin{array}{cc}R_{11} & R_{12} \\ 0 & 0\end{array}\right) .
\end{displaymath}">
</DIV>
<BR CLEAR="ALL">
<P></P>
Then <B><I>k</I></B> is the effective rank of <B><I>A</I></B>.
See Golub and Van Loan [<A
HREF="node151.html#GVL2">55</A>]
for a further discussion of numerical rank determination.
<A NAME="2704"></A><A NAME="2705"></A>
<P>
The so-called basic solution to the linear least squares
problem (<A HREF="node27.html#llsq">2.1</A>)<A NAME="2707"></A> can be obtained from this factorization as
<BR><P></P>
<DIV ALIGN="CENTER">
<!-- MATH
\begin{displaymath}
x = P \left( \begin{array}{c} R_{11}^{-1} \hat{c}_1 \\0 \end{array} \right),
\end{displaymath}
-->
<IMG
WIDTH="144" HEIGHT="54" BORDER="0"
SRC="img116.png"
ALT="\begin{displaymath}
x = P \left( \begin{array}{c} R_{11}^{-1} \hat{c}_1 \\ 0 \end{array} \right),
\end{displaymath}">
</DIV>
<BR CLEAR="ALL">
<P></P>
where <IMG
WIDTH="20" HEIGHT="32" ALIGN="MIDDLE" BORDER="0"
SRC="img117.png"
ALT="$\hat{c}_1$">
consists of just the first <B><I>k</I></B> elements of <B><I>c</I> = <I>Q</I><SUP><I>T</I></SUP> <I>b</I></B>.
<P>
The <B><I>QR</I></B> factorization with column pivoting can be computed either by
subroutine xGEQPF<A NAME="2715"></A><A NAME="2716"></A><A NAME="2717"></A><A NAME="2718"></A>
or by subroutine
xGEQP3.<A NAME="2719"></A><A NAME="2720"></A><A NAME="2721"></A><A NAME="2722"></A>
Both subroutines compute the factorization but do not attempt to
determine the rank of <B><I>A</I></B>. xGEQP3 is a Level 3 BLAS version of <B><I>QR</I></B> with
column pivoting and is considerably faster than xGEQPF, while maintaining
the same numerical behavior. The difference between the two routines
can best be described as follows. For each column, the subroutine xGEQPF
selects one column, permutes it, computes the reflector that zeroes some
of its components, and applies it to the rest of the matrix via Level 2
BLAS operations. The subroutine xGEQP3, however, only updates one column
and one row of the rest of the matrix (information necessary for the
next pivoting phase) and delays the update of the rest of the matrix
until a block of columns has been processed. This resulting block
of reflectors is then applied to the rest of the matrix as a Level 3 BLAS
operation. xGEQPF has been retained for compatibility with Release 2.0
of LAPACK, but we omit references to this routine in the remainder
of this users' guide.
<P>
For both subroutines, the matrix <B><I>Q</I></B> is represented in exactly the same way
as after a call of
xGEQRF<A NAME="2723"></A><A NAME="2724"></A><A NAME="2725"></A><A NAME="2726"></A>, and so the
routines xORGQR and xORMQR can be used to work with <B><I>Q</I></B>
(xUNGQR and xUNMQR if <B><I>Q</I></B> is complex).
<A NAME="2727"></A><A NAME="2728"></A><A NAME="2729"></A><A NAME="2730"></A>
<A NAME="2731"></A><A NAME="2732"></A><A NAME="2733"></A><A NAME="2734"></A>
<P>
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<ADDRESS>
<I>Susan Blackford</I>
<BR><I>1999-10-01</I>
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