Prove that $k(A) \geq \rho(A)/\min |\lambda|$ and that $k(A) \geq \rho(A) \rho(A^{-1})$

Question

I'm trying to solve this question here. Thank you in advance for your help.

Prove that $k(A) \geq \rho(A)/\min |\lambda|$ and that $k(A) \geq \rho(A) \rho(A^{-1}) $

We assume the matrices $A$ and $A^{-1}$ exist. Any size $A$ matrix.

I know $k(A) = ||A||\cdot ||A^{-1}||$ and $\rho(A) = \max_{\lambda\in\sigma(A)} |\lambda|=\lim_{n\to\infty}\|A^n\|^{1/n}$ with $\rho$ being the spectral radius and $k(A)$ being the condition number of the matrix.

I'm not sure what exactly I need to set it up in order to get to both needed results.

Answer 1

We will prove $k(A) \geq \rho(A)\rho(A^{-1}) $

We know, by definition, let $ \lambda $ be the eigenvalue for which $ \rho(A) = \max_{\lambda\in\sigma(A)} |\lambda| $. We let $x$ be an eigenvector, $ ||x||_{v} = 1 $.

$\Rightarrow \rho(A) = \max_{\lambda\in\sigma(A)} |\lambda| = ||\lambda x||_{v}$

$\Rightarrow ||\lambda x||_{v} \leq ||Ax||_{v} \leq ||A||\cdot||x||_{v} = ||A|| $

$\Rightarrow \rho(A) \leq ||A||$

We apply the same with $A^{-1}$:

$\Rightarrow \rho(A^{-1}) = \max_{\lambda\in\sigma(A^{-1})} |\lambda| = ||\lambda x||_{v}$

$\Rightarrow ||\lambda x||_{v} \leq ||A^{-1}x||_{v} \leq ||A^{-1}||\cdot||x||_{v} = ||A^{-1}|| $

$\Rightarrow \rho(A^{-1}) \leq ||A^{-1}||$

We combine both inequalities and get $k(A) \geq \rho(A)\rho(A^{-1})$.

To get $ k(A) \geq \rho(A)/\min_{\lambda\in\sigma(A)} |\lambda| $, we note $ k(A) \geq \rho(A)\rho(A^{-1})$. We know $\rho(A) = \max_{\lambda\in\sigma(A)} |\lambda|$ and the same with $\rho(A^{-1})$. We note that the eigenvalues of $A^{-1}$ are the reciprocals of those of $A$.

$\Rightarrow k(A)\geq \rho(A)\rho(A^{-1})$

$\Rightarrow k(A) \geq \rho(A)/\min_{\lambda\in\sigma(A)}|\lambda|$.