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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

lnormnorm
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  • Please include in your question what you have tried. eg have you computed $A^{-1}$ (in terms of $\epsilon$)? – stewbasic Oct 04 '16 at 05:25
  • Did you compute $A^{-1}$? – copper.hat Oct 04 '16 at 05:26
  • I get $\operatorname{cond}\infty T = \max(3,{2 \over |\epsilon|}) \max (2, 2 |\epsilon|)$, $\operatorname{cond}\infty T^T = (1+|\epsilon|)\max(2,{2 \over |\epsilon|} )$. – copper.hat Oct 04 '16 at 05:36
  • @copper.hat What method did you use to calculate the condition number of T? – lnormnorm Oct 04 '16 at 05:42
  • $|T|\infty |T^{-1}|\infty$. – copper.hat Oct 04 '16 at 05:43
  • So can we not reproduce cond(T) = 5? – lnormnorm Oct 04 '16 at 05:47
  • I made some computational mistakes above, but, regardless of norm chosen, the condition number of $T$ (and $T^T$) varies as ${1 \over \epsilon}$ for small $\epsilon$. (This is because all norms are equivalent.) – copper.hat Oct 04 '16 at 05:50
  • Note that $|A|_\infty$ is just the $\max$ row sum (of the absolute values), so is easy to compute. – copper.hat Oct 04 '16 at 05:57
  • +1 Is $\left\Vert |T^{-1}||T| \right\Vert_{\inf}$ the same thing as $\left\Vert T \right\Vert_{\inf} \left\Vert T^{-1} \right\Vert_{\inf}$ ? – lnormnorm Oct 04 '16 at 06:08
  • No, I presumed that was a typo., since we're talking about condition number. Usually, $\operatorname{cond} A = |A| |A^{-1}|$. However, I am frequently mistaken :-). – copper.hat Oct 04 '16 at 12:56
  • Do you have a more detailed description of Higham's paper, he seems to be a prolific lad, so a search turns up more than I am willing to troll through. – copper.hat Oct 05 '16 at 13:25
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    @copper.hat Higham [2002, pg144] Accuracy and Stability of Numerical Algorithms – lnormnorm Oct 09 '16 at 18:32
  • The condition number that Higham uses in that section is the one you have in the question which is different than the condition number one typically uses in the context of matrices. If you compute $|T^{-1}| |T|$ and take the $ \infty$ induced norm, then the result is $5$, as above. – copper.hat Oct 10 '16 at 02:24

1 Answers1

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

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

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

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

copper.hat
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