# Matrix and its condition number

This example is given in Higham , 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

• Please include in your question what you have tried. eg have you computed $A^{-1}$ (in terms of $\epsilon$)? – stewbasic Oct 4 '16 at 5:25
• Did you compute $A^{-1}$? – copper.hat Oct 4 '16 at 5:26
• 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 4 '16 at 5:36
• @copper.hat What method did you use to calculate the condition number of T? – lnormnorm Oct 4 '16 at 5:42
• @copper.hat Higham [2002, pg144] Accuracy and Stability of Numerical Algorithms – lnormnorm Oct 9 '16 at 18:32

## 1 Answer

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

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

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

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