For given matrix $A \in R^{n\times n}$, identity matrix $I$ and constant $c > 0$ is this possible to express $cond(A + cI)$ knowing $cond(A)$ and $c$?
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$\begingroup$ is this problem from a homework set? $\endgroup$– Aron AhmadiaMar 12, 2012 at 12:41
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$\begingroup$ @AronAhmadia No, it's from my head. I'm not student. Is it so obviously easy and stupid question? :) Since I could not answer it myself... I tried to derive it through SVD, then made some numerical tests and they don't match. $\endgroup$– AlexanderMar 12, 2012 at 12:46
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1$\begingroup$ Moderators. Maybe this question rather belongs to math.stackexchange.com Can you consider to move it there? $\endgroup$– AlexanderMar 12, 2012 at 15:39
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$\begingroup$ Given the importance of condition numbers for linear systems in numerical linear algebra, and that the answer by Professor Neumaier rightly corrects a common misconception about condition number, I'd rather leave the answer here for its educational value. $\endgroup$– Geoff OxberryMar 14, 2012 at 11:55
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2 Answers
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There is hardly any connection. First, the condition number depends on the norm used. Second, even in a fixed norm, the eigenvalues don't tell much about the condition number.
In the 2-norm, the condition number is the ration of the largest and the smallest singular value. For symmetric positive definite matrices, this becomes the ratio of the largest and smallest eigenvalue.
I leave it as an easy exercise that if $A$ is symmetric positive definite and you know the condition of $A$ and $A+c_0I$ for some $c_0>0$, you can calculate the condition number of all $A+cI$ with $c>0$.
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It depends on the actual eigenvalues, so no. Counter example:
$$ A_1 = \begin{bmatrix} 1&0\\ 0&2 \end{bmatrix}, \quad A_2 = \begin{bmatrix} 100&0\\ 0&200 \end{bmatrix} $$ Condition number for both matrices are 2, but when $cI$ is added, $\mbox{cond}(A_1 + cI) = \frac{2+c}{1+c}$ while $\mbox{cond}(A_2 + cI) = \frac{200+c}{100+c}$.
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$\begingroup$ Let's assume we have eigenvalues of the $A$. What is condition number of $A+cI$ then? $\endgroup$ Mar 12, 2012 at 20:07
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$\begingroup$ $\frac{\lambda_{max} + c}{\lambda_{min} + c}$, should be smaller if $c>0$. $\endgroup$– dranxoMar 13, 2012 at 0:04
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1$\begingroup$ Isn't this true ONLY for symmetric matrices? $\endgroup$– InquestMar 13, 2012 at 4:50
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$\begingroup$ Oops now I see. Yeah good point. Normally the condition number is defined in terms of the singular values. In the symmetric case eigenvalues are fine. $\endgroup$– dranxoMar 13, 2012 at 5:37