5
$\begingroup$

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 $c$?

$\endgroup$
4
  • $\begingroup$ is this problem from a homework set? $\endgroup$ Mar 12, 2012 at 12:41
  • $\begingroup$ @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. $\endgroup$
    – Alexander
    Mar 12, 2012 at 12:46
  • 1
    $\begingroup$ Moderators. Maybe this question rather belongs to math.stackexchange.com Can you consider to move it there? $\endgroup$
    – Alexander
    Mar 12, 2012 at 15:39
  • $\begingroup$ 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. $\endgroup$ Mar 14, 2012 at 11:55

2 Answers 2

5
$\begingroup$

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 $c>0$.

$\endgroup$
4
$\begingroup$

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 $\mbox{cond}(A_2 + cI) = \frac{200+c}{100+c}$.

$\endgroup$
5
  • $\begingroup$ Let's assume we have eigenvalues of the $A$. What is condition number of $A+cI$ then? $\endgroup$
    – Alexander
    Mar 12, 2012 at 20:07
  • $\begingroup$ $\frac{\lambda_{max} + c}{\lambda_{min} + c}$, should be smaller if $c>0$. $\endgroup$
    – dranxo
    Mar 13, 2012 at 0:04
  • 1
    $\begingroup$ Isn't this true ONLY for symmetric matrices? $\endgroup$
    – Inquest
    Mar 13, 2012 at 4:50
  • $\begingroup$ $(A+cI)v = (\lambda + c)v$? $\endgroup$
    – dranxo
    Mar 13, 2012 at 5:30
  • $\begingroup$ 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. $\endgroup$
    – dranxo
    Mar 13, 2012 at 5:37

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.