The Convergence Of SOR iteration method

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SUMMARY

The discussion focuses on the convergence of the Successive Over-Relaxation (SOR) iteration method for a specific system of equations. The participant initially concluded that the coefficient matrix is not positive definite, which is crucial for SOR convergence. However, upon further examination, it was revealed that the eigenvalues of the coefficient matrix are indeed positive, indicating that SOR will converge for any initial guess. The key condition for convergence is that the spectral radius ρ(Tω) must be less than 1, which was miscalculated initially.

PREREQUISITES
  • Understanding of the SOR iteration method
  • Knowledge of eigenvalues and eigenvectors
  • Familiarity with matrix norms
  • Basic concepts of linear algebra
NEXT STEPS
  • Study the properties of positive definite matrices
  • Learn how to compute the spectral radius of a matrix
  • Explore the implications of the relaxation factor ω in SOR
  • Investigate alternative iterative methods for solving linear systems
USEFUL FOR

Mathematicians, engineers, and students involved in numerical analysis or computational mathematics, particularly those working with iterative methods for solving linear equations.

sigh1342
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Homework Statement


show SOR iteration method converges for the system.
$$6x+4y+2z=11$$
$$4x+7y+4z=3$$
$$2x+4y+5=-3$$


Homework Equations



if the coeff. matrix is positive definite matrix and 0≤ω≤2. Then SOR converge for any initial guess.
Or if $$ρ(T_{ω})$$≥|ω-1|, then SOR converge for any initial guess.ρ(T) means the largest magnitude of all eigenvalue of T.$$T_{ω}=(I − ωL)^{-1} ((1 − ω)I + ωU)$$
Or any norm of $$T_{ω} <1 $$ Then SOR converge for any initial guess

The Attempt at a Solution


I found that the coeff. matrix is not positive definite matrix . and the ρ(T) is hard to find .
Any other method ? Or what I miss. Thanks you :blushing:
 
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sigh1342 said:
I found that the coeff. matrix is not positive definite matrix

Can you explain how you drew that conclusion? The coefficient matrix has positive eigenvalues.
 
fzero said:
Can you explain how you drew that conclusion? The coefficient matrix has positive eigenvalues.

Oh I find that the value $$x^tAx$$ that I was computed is wrong . :frown:
Thanks you so much :-p
 

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