I want to prove that the condition number of a positive definite matrix $S$ is
$$ k_2(S)= \frac{\max \lambda_i}{\min \lambda_i} $$
where $k_2$ is condition number for spectral norm. Can someone please help me with this?
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2 Answers
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A positive definite matrix is diagonalizable and invertible, meaning all of its eigenvalues are nonzero. Can you see how it works for a diagonal matrix? (The norm of a diagonal matrix is the absolute value of its greatest eigenvalue.)
answered Oct 16, 2020 at 13:47
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The condition number in any norm is given by $k = \| S\| \|S^{-1}\|$. If $S$ has eigenvalues $\lambda_1, \cdots, \lambda_n$, its inverse has eigenvalues $\frac{1}{\lambda_1}, \cdots \frac{1}{\lambda_n}$ and so, $ \|S\|_2 = \max|\lambda_i|, \quad \|S^{-1}\|_2 = \max |\frac{1}{\lambda_i}|=\frac{1}{\min|\lambda_i|}$. This leads to the result: $$ k_2(S) = \|S\|_2 \|S^{-1}\|_2 = \dfrac{\displaystyle\max_{i=1,\cdots, n}|\lambda_i|}{\displaystyle \min_{i=1, \cdots, n}|\lambda_i|} $$
answered Oct 16, 2020 at 13:44
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$\begingroup$ So it works for any matrix?and we can use being positive definite only to show that all eigen values are positive? $\endgroup$– M.PtOct 16, 2020 at 13:56
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$\begingroup$ @M.Pt It works for invertible matrices, like the positive definite matrices. $\endgroup$ Oct 16, 2020 at 16:10
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$\begingroup$ A comment before the downvote would have been more constructive... $\endgroup$ Oct 22, 2021 at 13:18