Convergence of iteration method - Relation between norm and eigenvalue

[mathmari]

Hey!

Let $G$ be the iteration matrix of an iteration method. So that the iteration method converges is the only condition that the spectral radius id less than $1$, $\rho (G)<1$, no matter what holds for the norms of $G$ ?
I mean if it holds that $\|G\|_{\infty}=3$ and $\rho (G)=0.3<1$ or $\|G\|_{1}=1$ and $\rho (G)=0.1<1$ in these both cases the iteration method converges, or not? (Wondering)

I have also an other question. Let $A$ be a symmetric matrix. If $\|A\|=0.01$ then there is an eigenvalue $\lambda$ with $|\lambda|\leq 0.01$, isn't it? We get that using the spectral radius: $\rho (A)=\max |\lambda_i|\leq \|A\|_1$, right? (Wondering)

 

[Opalg]

Let $G$ be the iteration matrix of an iteration method. So that the iteration method converges is the only condition that the spectral radius id less than $1$, $\rho (G)<1$, no matter what holds for the norms of $G$ ?
I mean if it holds that $\|G\|_{\infty}=3$ and $\rho (G)=0.3<1$ or $\|G\|_{1}=1$ and $\rho (G)=0.1<1$ in these both cases the iteration method converges, or not? (Wondering)

For an iteration procedure, you are going to be interested in powers of $G$. So what matters is that $\|G^n\| < 1$ when $n$ is large. It doesn't matter whether $G$ itself has norm larger than $1$, so long as the powers of $G$ have smaller norms.

The spectral radius has the property that $$\rho(G) = \lim_{n\to\infty}\|G^n\|^{1/n}.$$ So if $\rho(G)<1$ then $\|G^n\|<1$ whenever $n$ is large enough, and that is sufficient to ensure that the iteration method converges.

I have also an other question. Let $A$ be a symmetric matrix. If $\|A\|=0.01$ then there is an eigenvalue $\lambda$ with $|\lambda|\leq 0.01$, isn't it? We get that using the spectral radius: $\rho (A)=\max |\lambda_i|\leq \|A\|_1$, right? (Wondering)

For a symmetric matrix, the spectral radius is equal to the norm. So if $\|A\|=0.01$ then there is actually an eigenvalue $\lambda$ with $|\lambda| = 0.01$.

 

[mathmari]

For an iteration procedure, you are going to be interested in powers of $G$. So what matters is that $\|G^n\| < 1$ when $n$ is large. It doesn't matter whether $G$ itself has norm larger than $1$, so long as the powers of $G$ have smaller norms.

The spectral radius has the property that $$\rho(G) = \lim_{n\to\infty}\|G^n\|^{1/n}.$$ So if $\rho(G)<1$ then $\|G^n\|<1$ whenever $n$ is large enough, and that is sufficient to ensure that the iteration method converges.For a symmetric matrix, the spectral radius is equal to the norm. So if $\|A\|=0.01$ then there is actually an eigenvalue $\lambda$ with $|\lambda| = 0.01$.

I see! Thank you very much! (Smile)