MHB Convergence of iteration method - Relation between norm and eigenvalue

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The discussion focuses on the convergence of iteration methods using the iteration matrix G, emphasizing that the spectral radius, ρ(G), must be less than 1 for convergence, regardless of the norms of G. It is clarified that even if the norm of G exceeds 1, the powers of G can still converge as long as ρ(G) is less than 1. Additionally, for a symmetric matrix A, if the norm is 0.01, there exists an eigenvalue with an absolute value equal to 0.01, as the spectral radius is equal to the norm in this case. The relationship between the spectral radius and the norms is reinforced, highlighting its significance in determining convergence. Understanding these properties is crucial for analyzing iteration methods and eigenvalue behavior.
mathmari
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Hey! :o

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)
 
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mathmari said:
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.

mathmari said:
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 $|\lambda| = 0.01$.
 
Opalg said:
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)
 
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