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This example is given in Higham [2002], and is provided without explanation. I am not sure how the condition number of the matrix is just 5. How can you directly calculate the condition number of a matrix with epsilon? I know that cond(A) = $\left\Vert |A^{-1}||A| \right\Vert_{\inf}$, but I'm not sure how that helps here.

I calculated $A^{-1} = \{(1, -1/\epsilon, 1),(0, 1/\epsilon, -1), (0, 0, 1)\}$

Example

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  • $\begingroup$ Please include in your question what you have tried. eg have you computed $A^{-1}$ (in terms of $\epsilon$)? $\endgroup$ – stewbasic Oct 4 '16 at 5:25
  • $\begingroup$ Did you compute $A^{-1}$? $\endgroup$ – copper.hat Oct 4 '16 at 5:26
  • $\begingroup$ I get $\operatorname{cond}_\infty T = \max(3,{2 \over |\epsilon|}) \max (2, 2 |\epsilon|)$, $\operatorname{cond}_\infty T^T = (1+|\epsilon|)\max(2,{2 \over |\epsilon|} )$. $\endgroup$ – copper.hat Oct 4 '16 at 5:36
  • $\begingroup$ @copper.hat What method did you use to calculate the condition number of T? $\endgroup$ – lnormnorm Oct 4 '16 at 5:42
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    $\begingroup$ @copper.hat Higham [2002, pg144] Accuracy and Stability of Numerical Algorithms $\endgroup$ – lnormnorm Oct 9 '16 at 18:32
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$|T| = \begin{bmatrix} 1 & 1 & 0 \\ 0 & |\epsilon| & |\epsilon| \\ 0 & 0 & 1 \end{bmatrix} $

$|T^{-1}| = \begin{bmatrix} 1 & {1 \over |\epsilon|} & 1 \\ 0 & {1 \over |\epsilon|} & 1 \\ 0 & 0 & 1\end{bmatrix}$

$|T||T^{-1}| = \begin{bmatrix} 1 & 2 & 2 \\ 0 & 1 & 2 \\ 0 & 0 & 1 \end{bmatrix}$

$\| |T||T^{-1}| \|_\infty = 5$.

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