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

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.

• Your hint doesn't look right. – user10354138 Sep 29 '18 at 6:43
• @Brahadeesh It's any size matrix $A$. Just a general matrix. The same with $\lambda$ and $\rho$. It's using the definitions of them to obtain the inequalities. – MelMarieHA Sep 29 '18 at 7:19
• @user10354138 $\rho$ is the spectral radius, and I double checked my hint/note and that is correct. – MelMarieHA Sep 29 '18 at 7:20
• your hint is $k(A)=\rho(A)\rho(A^{-1})$ is definitely not correct, otherwise you wouldn't be asked to prove $k(A)\geq\rho(A)\rho(A^{-1})$. – user10354138 Sep 29 '18 at 7:37
• @user10354138 Very true! I'll remove it then! Does it make more sense? – MelMarieHA Sep 29 '18 at 7:53

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 $$A$$.

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

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