# Upper bound condition number

Let $$A$$ be a real matrix with all eigenvalues in the interval $$(0,1)$$. Show that $$\kappa(A)\le\dfrac{2-\lambda_{\min}(A)}{\lambda_{\min}(A)}$$ where $$\kappa(A)$$ is $$A$$'s condition number (maximum singular value $$\sigma_{\max}(A)$$ divided by minimum singular value $$\sigma_{\min}(A)$$) and $$\lambda_{\min}(A)$$ is $$A$$'s smallest eigenvalue. Inspired by this if we also assume that $$N$$ is positive definite.

Let $$\lambda_{\max}(A)$$ be $$A$$'s biggest eigenvalue. Then $$\dfrac{\lambda_{\max}(A)}{\lambda_{\min}(A)}<\dfrac{1}{\lambda_{\min(A)}}.$$ $$\sigma_{\max}(A)\ge\lambda_{\max}(A)$$ and $$\sigma_{\min}(A)\le\lambda_{\min}(A)$$, so $$\dfrac{\lambda_{\max}(A)}{\lambda_{\min}(A)}\le\kappa(A).$$ If $$A$$ is normal, equality holds and we're done. I don't know how to proceed if $$A$$ isn't normal.

• What is $N$? ${}{}$ Dec 12, 2019 at 17:48
• It's only in the link. $A=M-N$. Dec 12, 2019 at 17:56
• A problem should be reasonably self contained. Dec 12, 2019 at 17:56

Let $$A=\begin{bmatrix} {1 \over 2} & 10,000 \\ 0 & {1 \over 2} \end{bmatrix}$$.
The above formula suggests that an upper bound for the condition number is $$3$$, but $$\|A^{-1} e_1 \| = 2$$ and $$\| Ae_2 \| > 10,000$$ so we have $$\kappa(A) > 20,000$$.
• Thanks. Is the link wrong or does $M^{-1}A$ satisfy something I didn't capture in my reformulation? Dec 12, 2019 at 17:59