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I am trying to prove that $$\mbox{cond}_2(A)\ =\ \inf_{E\in\mathscr{E}}\mbox{cond}_2(E),\;\;\; \mbox{ where }\;\; A = \left(\begin{array}{cc} 100 & 99\\99 & 98 \end{array}\right).$$ And $\mathscr{E}$ is the set of matrices of order 2, whose elements $a_{ij}$ are integers satisfying $0\leq a_{ij}\leq 100$.

In order to prove it, it's clear that $\mbox{cond}_2(A) \geq \inf\limits_{E\in\mathscr{E}}\mbox{cond}_2(E)$, and I have showed that for every general $A$ of order 2, that $$\mbox{cond}_2(A)\ =\ \sigma + \sqrt{\sigma^2-1},\;\;\; \mbox{ with }\;\; \sigma = \frac{\sum\limits_{i,j=1}^2|a_{ij}|^2}{2|\det(A)|},$$ but I don't know how to prove that $\mbox{cond}_2(A) \leq \inf\limits_{E\in\mathscr{E}}\mbox{cond}_2(E)$ for that particular $A$. Please, somebody help me. Thanks in advance.

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  • $\begingroup$ It doesn't sound right: the condition number of the identity matrix is $1$, but the condition number of your $A$ is clearly much larger. Are you sure you have jotted down the problem statement correctly? $\endgroup$
    – user1551
    Apr 13, 2013 at 17:12
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    $\begingroup$ Yes, its the problem 2.2-4 of the book: Numerical Linear Algebra and Optimisation by P. Ciarlet. But you are right, that problem is wrong. Thanks you $\endgroup$
    – user70195
    Apr 13, 2013 at 17:25
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    $\begingroup$ I know what is the problem, the problem say "inf", but the correct statement must be: $$\mbox{cond}_2(A)\ =\ \sup_{E\in\mathscr{E}}\mbox{cond}_2(E).$$ $\endgroup$
    – FASCH
    Apr 13, 2013 at 19:48

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The correct statement should be that $\operatorname{cond}_2(A)=\sup\{\operatorname{cond}_2(A):E\in\mathscr{E} \text{ and } E \text{ is invertible}\}$. First, note that the condition number is an increasing function of your $\sigma$. The denominatior of $\sigma$ is twice the absolute value of the determinant. Since $\det(A)=-1$, the given $A$ attains the least possible determinant among all invertible integer matrices. So it remains to show that if the numerator of $\sigma$ of some $E\neq A$ (i.e. $\|E\|_F^2$) is greater than or equal to $\|A\|_F^2$, then either $E$ is singular or $\operatorname{cond}_2(E)\le\operatorname{cond}_2(A)$. Since there are only a few matrices whose squared Frobenius norms are greater than or equal to $\|A\|_F^2$, you only need to do check their condition numbers exhaustively.

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