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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.

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

1 Answer 1

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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 $\kappa(A) > 20,000$.

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  • $\begingroup$ Thanks. Is the link wrong or does $M^{-1}A$ satisfy something I didn't capture in my reformulation? $\endgroup$
    – herokenzan
    Commented Dec 12, 2019 at 17:59
  • $\begingroup$ I did not look at the link. I am just pointing out that the formula in the question is not true in general. $\endgroup$
    – copper.hat
    Commented Dec 12, 2019 at 18:00

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