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Given a $n \times n$ matrix $A$ with condition number $1$, prove or disprove $\mbox{cond} (A) = \mbox{cond} (PA)$, where $P$ is is any permutation matrix.

My attempt:
$cond(A) = ||A||*||A^{-1}||$
$cond(PA) = ||PA||*||(PA)^{-1}||$

(From here I am not sure I am doing it correctly:)

From the definition of the matrix norm:
${\displaystyle \left\|A\right\|_{2}\leq \left(\sum _{i=1}^{m}\sum _{j=1}^{n}|a_{ij}|^{2}\right)^{1/2}=\left\|A\right\|_{F},} $

It clearly follows that ||PA|| = ||A||, since P has only the effect of permuting the values inside A without changing them.

Similarly, $||(PA)^{-1}|| = ||A^{-1}||.$

Thus, cond(A) = cond(PA)

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  • $\begingroup$ Yes this is true $\endgroup$ Commented Oct 11, 2016 at 8:17

1 Answer 1

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If $\mathrm P$ is a permutation matrix, then $\mathrm P^{-1} = \mathrm P^T$. Hence,

$$\kappa (\mathrm P \mathrm A) = \dfrac{\sigma_{\max} (\mathrm P \mathrm A)}{\sigma_{\min} (\mathrm P \mathrm A)} = \sqrt{\dfrac{\lambda_{\max} (\mathrm A^T \mathrm P^T \mathrm P \mathrm A)}{\lambda_{\min} (\mathrm A^T \mathrm P^T \mathrm P \mathrm A)}} = \sqrt{\dfrac{\lambda_{\max} (\mathrm A^T \mathrm A)}{\lambda_{\min} (\mathrm A^T \mathrm A)}} = \dfrac{\sigma_{\max} (\mathrm A)}{\sigma_{\min} (\mathrm A)} = \kappa (\mathrm A)$$

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