# Condition number of a positive definite matrix

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?

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.)

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: $$k_2(S) = \|S\|_2 \|S^{-1}\|_2 = \dfrac{\displaystyle\max_{i=1,\cdots, n}|\lambda_i|}{\displaystyle \min_{i=1, \cdots, n}|\lambda_i|}$$

• So it works for any matrix?and we can use being positive definite only to show that all eigen values are positive?
– M.Pt
Oct 16, 2020 at 13:56
• @M.Pt It works for invertible matrices, like the positive definite matrices. Oct 16, 2020 at 16:10
• A comment before the downvote would have been more constructive... Oct 22, 2021 at 13:18