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Further Details: Error Bounds for Linear Equality Constrained Least Squares Problems

In this subsection, we will summarize the available error bound. The reader may also refer to [2,13,18,50] for further details.

Let $\widehat {x}$ be the solution computed by the driver xGGLSE (see subsection 4.6). It is normwise stable in a mixed forward/backward sense [18,13]. Specifically, $\widehat {x} = \bar {x} + \Delta \bar {x}$, where $\bar {x}$ solves $\min\{\Vert b + \Delta b - (A + \Delta A)x \Vert _2: \; (B + \Delta B)x = d \} $, and

$\textstyle \parbox{2in}{
\begin{eqnarray*}
\Vert \Delta \bar {x} \Vert _2 & \...
... \Delta A \Vert _F & \leq & q(m,n,p)\epsilon\Vert A\Vert _F,
\end{eqnarray*} }$     $\textstyle \parbox{2in}{
\begin{eqnarray*}
\Vert \Delta b \Vert _2 & \leq & q...
...t \Delta B \Vert _F & \leq & q(m,n,p)\epsilon\Vert B\Vert _F,
\end{eqnarray*}}$
q(m,n,p) is a modestly growing function of m, n, and p. We take q(m,n,p) = 1 in the code fragment above. Let $X^{\dagger}$ denote the Moore-Penrose pseudo-inverse of X. Let $\kappa_B(A) = \Vert A \Vert _F \Vert (AP)^\dagger \Vert _2 $( = CNDAB above) and $\kappa_A(B) = \Vert B \Vert _F \Vert B^\dagger_A \Vert _2 $( = CNDBA above) where $P = I - B^\dagger B$ and $B^\dagger_A = (I - (AP)^\dagger A)B^\dagger$. When $q(m,n,p)\epsilon$ is small, the error $x-\hat{x}$ is bounded by

\begin{displaymath}
% latex2html id marker 15239\frac{ \Vert x-\hat{x} \Vert ...
...rt r \Vert _2 }{ \Vert A \Vert _F \Vert x \Vert _2 } \right\}.
\end{displaymath}



When B = 0 and d = 0, we essentially recover error bounds for the linear least squares (LS) problem:

\begin{displaymath}
\frac{ \Vert x-\widehat {x} \Vert _2 }{ \Vert x \Vert _2 } ...
...rt _2 }{ \Vert A \Vert _F \Vert x \Vert _2 } \right) \right\},
\end{displaymath}

where $\kappa(A) = \Vert A \Vert _F \Vert A^\dagger \Vert _2 $. Note that the error in the standard least squares problem provided in section 4.5.1 is

\begin{displaymath}
% latex2html id marker 13969\frac{ \Vert x-\widehat {x} \...
...^2(A)\frac{ \Vert r \Vert _2 }{ \Vert Ax \Vert _2 }\right\}\\
\end{displaymath}

since $\sin(\theta) = \frac{ \Vert A\widehat {x}-b \Vert _2 }{ \Vert b \Vert _2 }$. If one assumes that q(m,n) = p(n) = 1, then the bounds are essentially the same.


next up previous contents index
Next: General Linear Model Problem Up: Linear Equality Constrained Least Previous: Linear Equality Constrained Least   Contents   Index
Susan Blackford
1999-10-01