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<H2><A NAME="SECTION034121000000000000000">
Further Details: Error Bounds for the Generalized Singular Value Decomposition</A>
</H2>
<P>
The GSVD algorithm used in LAPACK ([<A
HREF="node151.html#paige86a">83</A>,<A
HREF="node151.html#baidemmel92b">10</A>,<A
HREF="node151.html#baizha93">8</A>])
is backward stable:
<A NAME="12964"></A>
<A NAME="12965"></A>
<A NAME="12966"></A>
<P>
<P>
<BLOCKQUOTE>Let the computed GSVD of <B><I>A</I></B> and <B><I>B</I></B> be
<!-- MATH
$\hat{U} \hat{\Sigma}_1 [0,\hat{R}] \hat{Q}^T$
-->
<IMG
WIDTH="102" HEIGHT="41" ALIGN="MIDDLE" BORDER="0"
SRC="img878.png"
ALT="$\hat{U} \hat{\Sigma}_1 [0,\hat{R}] \hat{Q}^T$">
and
<!-- MATH
$\hat{V} \hat{\Sigma}_2 [0,\hat{R}] \hat{Q}^T$
-->
<IMG
WIDTH="102" HEIGHT="41" ALIGN="MIDDLE" BORDER="0"
SRC="img879.png"
ALT="$\hat{V} \hat{\Sigma}_2 [0,\hat{R}] \hat{Q}^T$">.
This is nearly the exact GSVD of
<B><I>A</I>+<I>E</I></B> and <B><I>B</I>+<I>F</I></B> in the following sense. <B><I>E</I></B> and <B><I>F</I></B> are small:
</BLOCKQUOTE>
<BR><P></P>
<DIV ALIGN="CENTER">
<!-- MATH
\begin{displaymath}
\|E\|_2 / \|A\|_2 \leq p(n) \epsilon \; \; {\rm and} \; \;
\|F\|_2 / \|B\|_2 \leq p(n) \epsilon \; ;
\end{displaymath}
-->
<IMG
WIDTH="352" HEIGHT="31" BORDER="0"
SRC="img880.png"
ALT="\begin{displaymath}
\Vert E\Vert _2 / \Vert A\Vert _2 \leq p(n) \epsilon \; \; {...
...; \;
\Vert F\Vert _2 / \Vert B\Vert _2 \leq p(n) \epsilon \; ;
\end{displaymath}">
</DIV>
<BR CLEAR="ALL">
<P></P><BLOCKQUOTE>
there exist small
<!-- MATH
$\delta \hat{Q}$
-->
<IMG
WIDTH="27" HEIGHT="41" ALIGN="MIDDLE" BORDER="0"
SRC="img881.png"
ALT="$\delta \hat{Q}$">,
<!-- MATH
$\delta \hat{U}$
-->
<IMG
WIDTH="27" HEIGHT="21" ALIGN="BOTTOM" BORDER="0"
SRC="img882.png"
ALT="$\delta \hat{U}$">,
and
<!-- MATH
$\delta \hat{V}$
-->
<IMG
WIDTH="27" HEIGHT="21" ALIGN="BOTTOM" BORDER="0"
SRC="img883.png"
ALT="$\delta \hat{V}$">
such that
<!-- MATH
$\hat{Q} + \delta \hat{Q}$
-->
<IMG
WIDTH="62" HEIGHT="41" ALIGN="MIDDLE" BORDER="0"
SRC="img884.png"
ALT="$\hat{Q} + \delta \hat{Q}$">,
<!-- MATH
$\hat{U} + \delta \hat{U}$
-->
<IMG
WIDTH="62" HEIGHT="41" ALIGN="MIDDLE" BORDER="0"
SRC="img652.png"
ALT="$\hat{U}+ \delta \hat{U}$">,
and
<!-- MATH
$\hat{V} + \delta \hat{V}$
-->
<IMG
WIDTH="62" HEIGHT="41" ALIGN="MIDDLE" BORDER="0"
SRC="img653.png"
ALT="$\hat{V}+ \delta
\hat{V}$">
are exactly orthogonal (or unitary):
</BLOCKQUOTE>
<BR><P></P>
<DIV ALIGN="CENTER">
<!-- MATH
\begin{displaymath}
\| \delta \hat{Q} \|_2 \leq p(n) \epsilon \; \; , \; \;
\| \delta \hat{U} \|_2 \leq p(n) \epsilon \; \; {\rm and} \; \;
\| \delta \hat{V} \|_2 \leq p(n) \epsilon \; \; ;
\end{displaymath}
-->
<IMG
WIDTH="417" HEIGHT="31" BORDER="0"
SRC="img885.png"
ALT="\begin{displaymath}
\Vert \delta \hat{Q} \Vert _2 \leq p(n) \epsilon \; \; , \; ...
...\; \;
\Vert \delta \hat{V} \Vert _2 \leq p(n) \epsilon \; \; ;
\end{displaymath}">
</DIV>
<BR CLEAR="ALL">
<P></P><BLOCKQUOTE>
and
</BLOCKQUOTE>
<BR><P></P>
<DIV ALIGN="CENTER">
<!-- MATH
\begin{displaymath}
A + E = ( \hat{U} + \delta \hat{U} ) \hat{\Sigma}_1 [ 0, \hat{R}] (\hat{Q} + \delta \hat{Q})^T
\; \; {\rm and} \; \;
B + F = ( \hat{V} + \delta \hat{V} ) \hat{\Sigma}_2 [ 0, \hat{R}] (\hat{Q} + \delta \hat{Q})^T
\end{displaymath}
-->
<IMG
WIDTH="568" HEIGHT="31" BORDER="0"
SRC="img886.png"
ALT="\begin{displaymath}
A + E = ( \hat{U} + \delta \hat{U} ) \hat{\Sigma}_1 [ 0, \ha...
...V} ) \hat{\Sigma}_2 [ 0, \hat{R}] (\hat{Q} + \delta \hat{Q})^T
\end{displaymath}">
</DIV>
<BR CLEAR="ALL">
<P></P><BLOCKQUOTE>
is the exact GSVD of <B><I>A</I>+<I>E</I></B> and <B><I>B</I>+<I>F</I></B>. Here <B><I>p</I>(<I>n</I>)</B> is a modestly growing function of <B><I>n</I></B>, and
we take <B><I>p</I>(<I>n</I>)=1</B> in the above code fragment.
</BLOCKQUOTE>
<P>
<BLOCKQUOTE>Let <IMG
WIDTH="21" HEIGHT="30" ALIGN="MIDDLE" BORDER="0"
SRC="img72.png"
ALT="$\alpha_i$">
and <IMG
WIDTH="20" HEIGHT="32" ALIGN="MIDDLE" BORDER="0"
SRC="img73.png"
ALT="$\beta_i$">
be the square roots of the diagonal entries of the exact
<!-- MATH
$\Sigma_1^T \Sigma_1$
-->
<IMG
WIDTH="48" HEIGHT="38" ALIGN="MIDDLE" BORDER="0"
SRC="img887.png"
ALT="$\Sigma_1^T \Sigma_1$">
and
<!-- MATH
$\Sigma_2^T \Sigma_2$
-->
<IMG
WIDTH="48" HEIGHT="38" ALIGN="MIDDLE" BORDER="0"
SRC="img888.png"
ALT="$\Sigma_2^T \Sigma_2$">,
and let
<!-- MATH
$\hat{\alpha}_i$
-->
<IMG
WIDTH="21" HEIGHT="32" ALIGN="MIDDLE" BORDER="0"
SRC="img867.png"
ALT="$\hat{\alpha}_i$">
and <IMG
WIDTH="20" HEIGHT="41" ALIGN="MIDDLE" BORDER="0"
SRC="img868.png"
ALT="$\hat{\beta}_i$">
the square roots of the diagonal entries
of the computed
<!-- MATH
$\hat{\Sigma}_1^T \hat{\Sigma}_1$
-->
<IMG
WIDTH="48" HEIGHT="41" ALIGN="MIDDLE" BORDER="0"
SRC="img889.png"
ALT="$\hat{\Sigma}_1^T \hat{\Sigma}_1$">
and
<!-- MATH
$\hat{\Sigma}_2^T \hat{\Sigma}_2$
-->
<IMG
WIDTH="48" HEIGHT="41" ALIGN="MIDDLE" BORDER="0"
SRC="img890.png"
ALT="$\hat{\Sigma}_2^T \hat{\Sigma}_2$">.
Let
</BLOCKQUOTE>
<BR><P></P>
<DIV ALIGN="CENTER">
<!-- MATH
\begin{displaymath}
G = \left( \begin{array}{c} A \\B \end{array} \right) \; \; {\rm and} \; \; \hat{G} = \left( \begin{array}{c} \hat{A} \\\hat{B} \end{array} \right) \; .
\end{displaymath}
-->
<IMG
WIDTH="245" HEIGHT="54" BORDER="0"
SRC="img891.png"
ALT="\begin{displaymath}
G = \left( \begin{array}{c} A \\ B \end{array} \right) \; \;...
...( \begin{array}{c} \hat{A} \\ \hat{B} \end{array} \right) \; .
\end{displaymath}">
</DIV>
<BR CLEAR="ALL">
<P></P><BLOCKQUOTE>
Then provided <B><I>G</I></B> and <IMG
WIDTH="18" HEIGHT="21" ALIGN="BOTTOM" BORDER="0"
SRC="img892.png"
ALT="$\hat{G}$">
have full rank <B><I>n</I></B>, one can show [<A
HREF="node151.html#sun83">96</A>,<A
HREF="node151.html#paige84">82</A>] that
</BLOCKQUOTE>
<BR><P></P>
<DIV ALIGN="CENTER">
<!-- MATH
\begin{displaymath}
\left( \sum_{i=1}^n [( \hat{\alpha}_i - \alpha_i )^2 +
( \hat{\beta}_i - \beta_i )^2 ] \right)^{1/2} \leq
\frac{\sqrt{2} \left\| \left( \begin{array}{c} E \\F \end{array} \right) \right\|}
{\min ( \sigma_{\min} (G) , \sigma_{\min} (\hat{G}) )} \; \; .
\end{displaymath}
-->
<IMG
WIDTH="458" HEIGHT="82" BORDER="0"
SRC="img893.png"
ALT="\begin{displaymath}
\left( \sum_{i=1}^n [( \hat{\alpha}_i - \alpha_i )^2 +
( \h...
...{\min ( \sigma_{\min} (G) , \sigma_{\min} (\hat{G}) )} \; \; .
\end{displaymath}">
</DIV>
<BR CLEAR="ALL">
<P></P><BLOCKQUOTE>
In the code fragment we approximate the numerator of the last expression by
<!-- MATH
$\epsilon \|\hat{R}\|_{\infty}$
-->
<IMG
WIDTH="57" HEIGHT="41" ALIGN="MIDDLE" BORDER="0"
SRC="img894.png"
ALT="$\epsilon \Vert\hat{R}\Vert _{\infty}$">
and approximate the denominator by
<!-- MATH
$\| \hat{R}^{-1} \|^{-1}_{\infty}$
-->
<IMG
WIDTH="71" HEIGHT="41" ALIGN="MIDDLE" BORDER="0"
SRC="img895.png"
ALT="$\Vert \hat{R}^{-1} \Vert^{-1}_{\infty}$">
in order to compute <TT>SERRBD</TT>;
STRCON returns an approximation <TT>RCOND</TT> to
<!-- MATH
$1/ (\| \hat{R}^{-1} \|_{\infty} \| \hat{R} \|_{\infty})$
-->
<IMG
WIDTH="144" HEIGHT="41" ALIGN="MIDDLE" BORDER="0"
SRC="img896.png"
ALT="$1/ (\Vert \hat{R}^{-1} \Vert _{\infty} \Vert \hat{R} \Vert _{\infty})$">.
</BLOCKQUOTE>
<A NAME="13040"></A>
<P>
We assume that the rank <B><I>r</I></B> of <B><I>G</I></B> equals <B><I>n</I></B>, because otherwise the
<IMG
WIDTH="21" HEIGHT="30" ALIGN="MIDDLE" BORDER="0"
SRC="img72.png"
ALT="$\alpha_i$">s and <IMG
WIDTH="20" HEIGHT="32" ALIGN="MIDDLE" BORDER="0"
SRC="img73.png"
ALT="$\beta_i$">s are not well determined. For example, if
<BR><P></P>
<DIV ALIGN="CENTER">
<!-- MATH
\begin{displaymath}
A = \left( \begin{array}{cc} 10^{-16} & 0 \\0 & 1 \end{array} \right)
\; \; , \; \;
B = \left( \begin{array}{cc} 0 & 0 \\0 & 1 \end{array} \right)
\; \; {\rm and} \; \;
A' = \left( \begin{array}{cc} 0 & 0 \\0 & 1 \end{array} \right)
\; \; , \; \;
B' = \left( \begin{array}{cc} 10^{-16} & 0 \\0 & 1 \end{array} \right)
\end{displaymath}
-->
<IMG
WIDTH="627" HEIGHT="54" BORDER="0"
SRC="img897.png"
ALT="\begin{displaymath}
A = \left( \begin{array}{cc} 10^{-16} & 0 \\ 0 & 1 \end{arra...
...( \begin{array}{cc} 10^{-16} & 0 \\ 0 & 1 \end{array} \right)
\end{displaymath}">
</DIV>
<BR CLEAR="ALL">
<P></P>
then <B><I>A</I></B> and <B><I>B</I></B> have
<!-- MATH
$\alpha_{1,2} = 1, .7071$
-->
<IMG
WIDTH="113" HEIGHT="30" ALIGN="MIDDLE" BORDER="0"
SRC="img898.png"
ALT="$\alpha_{1,2} = 1, .7071$">
and
<!-- MATH
$\beta_{1,2} = 0, .7071$
-->
<IMG
WIDTH="112" HEIGHT="32" ALIGN="MIDDLE" BORDER="0"
SRC="img899.png"
ALT="$\beta_{1,2} = 0, .7071$">,
whereas
<B><I>A</I>'</B> and <B><I>B</I>'</B> have
<!-- MATH
$\alpha'_{1,2} = 0, .7071$
-->
<IMG
WIDTH="113" HEIGHT="34" ALIGN="MIDDLE" BORDER="0"
SRC="img900.png"
ALT="$\alpha'_{1,2} = 0, .7071$">
and
<!-- MATH
$\beta'_{1,2} = 1, .7071$
-->
<IMG
WIDTH="112" HEIGHT="34" ALIGN="MIDDLE" BORDER="0"
SRC="img901.png"
ALT="$\beta'_{1,2} = 1, .7071$">,
which
are completely different, even though
<!-- MATH
$\|A - A' \| = 10^{-16}$
-->
<IMG
WIDTH="139" HEIGHT="37" ALIGN="MIDDLE" BORDER="0"
SRC="img902.png"
ALT="$\Vert A - A' \Vert = 10^{-16}$">
and
<!-- MATH
$\|B - B' \| = 10^{-16}$
-->
<IMG
WIDTH="141" HEIGHT="37" ALIGN="MIDDLE" BORDER="0"
SRC="img903.png"
ALT="$\Vert B - B' \Vert = 10^{-16}$">.
In this case,
<!-- MATH
$\sigma_{\min} (G) = 10^{-16}$
-->
<IMG
WIDTH="131" HEIGHT="37" ALIGN="MIDDLE" BORDER="0"
SRC="img904.png"
ALT="$\sigma_{\min} (G) = 10^{-16}$">,
so <B><I>G</I></B> is nearly rank-deficient.
<P>
The reason the code fragment assumes <IMG
WIDTH="53" HEIGHT="30" ALIGN="MIDDLE" BORDER="0"
SRC="img105.png"
ALT="$m \geq n$">
is that in this case <IMG
WIDTH="18" HEIGHT="21" ALIGN="BOTTOM" BORDER="0"
SRC="img905.png"
ALT="$\hat{R}$">
is
stored overwritten on <B><I>A</I></B>, and can be passed to STRCON in order to compute
<TT>RCOND</TT>. If <IMG
WIDTH="53" HEIGHT="30" ALIGN="MIDDLE" BORDER="0"
SRC="img906.png"
ALT="$m \leq n$">,
then the
first <B><I>m</I></B> rows of <IMG
WIDTH="18" HEIGHT="21" ALIGN="BOTTOM" BORDER="0"
SRC="img905.png"
ALT="$\hat{R}$">
are
stored in <B><I>A</I></B>, and the last <B><I>n</I>-<I>m</I></B> rows of <IMG
WIDTH="18" HEIGHT="21" ALIGN="BOTTOM" BORDER="0"
SRC="img905.png"
ALT="$\hat{R}$">
are stored in <B><I>B</I></B>. This
complicates the computation of <TT>RCOND</TT>: either <IMG
WIDTH="18" HEIGHT="21" ALIGN="BOTTOM" BORDER="0"
SRC="img905.png"
ALT="$\hat{R}$">
must be copied to
a single array before calling STRCON, or else the lower level subroutine SLACON
must be used with code capable of solving linear equations with <IMG
WIDTH="18" HEIGHT="21" ALIGN="BOTTOM" BORDER="0"
SRC="img905.png"
ALT="$\hat{R}$">
and <IMG
WIDTH="28" HEIGHT="21" ALIGN="BOTTOM" BORDER="0"
SRC="img907.png"
ALT="$\hat{R}^T$">
as coefficient matrices.
<P>
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<ADDRESS>
<I>Susan Blackford</I>
<BR><I>1999-10-01</I>
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