A question we will run into later in the course asks how accurate we can expect the solution of a linear system to be if the right-hand side of the system has error in it.
Formally, this can be stated as follows: We wish to solve Ax=b, where AβCmΓm but the right-hand side has been perturbed by a small vector so that it becomes b+Ξ΄b.
Notice how the Ξ΄ touches the b. This is meant to convey that this is a symbol that represents a vector rather than the vector b that is multiplied by a scalar Ξ΄.
We would like to determine a formula, ΞΊ(A,b,Ξ΄b), that gives us a bound on how much a relative error in b is potentially amplified into a relative error in the solution x:
We assume that A has an inverse since otherwise there may be no solution or there may be an infinite number of solutions. To find an expression for \kappa( A, b, \delta\!b ) \text{,} we notice that
\begin{equation*}
\begin{array}{r c l}
A x + A \delta\!x \amp = \amp b + \delta\!b \\
A x \phantom{+ A \delta\!x } \amp = \amp b \hspace{0.25in} -\\ \hline
A \delta\!x \amp = \amp \phantom{b} \phantom{+} \delta\!b
\end{array}
\end{equation*}
A question becomes whether this is a pessimistic result or whether there are examples of b and \delta\!b for which the relative error in b is amplified by exactly \kappa( A ) \text{.} The answer is that, unfortunately, the bound is tight.
There is an {\widehat x} for which
\begin{equation*}
\| A \| = \max_{\| x \| = 1} \| A x \| = \| A {\widehat x} \|,
\end{equation*}
namely the x for which the maximum is attained. This is the direction of maximal magnification. Pick \widehat b = A {\widehat x} \text{.}
There is an \widehat {\delta\!b} for which
\begin{equation*}
\| A^{-1} \| = \max_{\| x \| \neq 0} \frac{\| A^{-1} x \|}{\| x \|}
=
\frac{\| A^{-1} \widehat{\delta\!b} \|}{\| \widehat{\delta\!b} \|},
\end{equation*}
This last exercise shows that there will always be choices for b and \delta\!b for which the relative error is at best directly translated into an equal relative error in the solution (if \kappa(
A ) = 1 ).