Which matrix is strictly diagonally dominant?
Iterative Methods for Linear Systems
Why iterative methods for large sparse systems, convergence and diagonal dominance, the Jacobi and Gauss–Seidel methods with worked examples, and successive over-relaxation (SOR).
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All 25 slides
- 1 Slide 1
- 2 Solving linear systems
- 3 Why to use iterative methods?
- 4 What are iterative methods?
- 5 What are iterative methods?
- 6 What are iterative methods?
- 7 What are iterative methods?
- 8 What are iterative methods?
- 9 What are iterative methods?
- 10 What are iterative methods?
- 11 What is convergence?
- 12 What is a diagonally dominant matrix?
- 13 Example:
- 14 Popular iterative methods
- 15 Jacobi method
- 16 Jacobi method (example)
- 17 Jacobi method (example)
- 18 Jacobi method (example)
- 19 Gauss-Seidel method
- 20 Gauss-Seidel method
- 21 Gauss-Seidel method (example)
- 22 Gauss-Seidel method (example 2)
- 23 Successive over relaxation method
- 24 Successive over relaxation method
- 25 Beyond basic methods
Why to use iterative methods?
Notes
Iterative methods start from a guess and improve it; good for large sparse systems where elimination is too costly.
What is a diagonally dominant matrix?
Notes
Strictly diagonally dominant: |aᵢᵢ| > Σ_{j≠i} |aᵢⱼ| in every row. Then Jacobi and Gauss–Seidel converge for any starting vector.
Exam angle
Jacobi method
Notes
Jacobi: xᵢ⁽ᵏ⁺¹⁾ = (bᵢ − Σ_{j≠i} aᵢⱼxⱼ⁽ᵏ⁾)/aᵢᵢ: every component uses the OLD vector.
Gauss-Seidel method
Notes
Gauss–Seidel: same formula but uses each NEW component as soon as it is computed; usually converges faster.
Memory trick Jacobi = all old values; Gauss–Seidel = newest values immediately.
Exam angle
One Jacobi step for 4x − y = 3, x + 5y = 6 from (0, 0).
Compare with the model answer
x = (3 + y)/4 = 0.75, y = (6 − x)/5 = 1.2 using the old x = 0, so (0.75, 1.2). (Gauss–Seidel would use x = 0.75: y = (6 − 0.75)/5 = 1.05.)