--- title: "Power method: tridiagonal 2,-1" subject: "Numerical Method" chapter: 3 key: "semester_4/numerical_method:power-method-tridiagonal-2-1" source: https://www.bctengineer.com/q/semester_4/numerical_method/power-method-tridiagonal-2-1 site: BCT Engineering --- # Power method: tridiagonal 2,-1 Numerical Method · Chapter 3: Solution of system of linear algebraic equations - Asked in: 2075 Baisakh, 8 marks; 2068 Bhadra, 8 marks (asked 2 times) - Answer checked by Claude Opus 5.5 (high effort) on 2026-10-14 - Page: https://www.bctengineer.com/q/semester_4/numerical_method/power-method-tridiagonal-2-1 - Whole chapter: https://www.bctengineer.com/md/semester_4/numerical_method/chapter-3 ## Question Find the dominant eigenvalue and eigenvector of the matrix: [2 -1 0; -1 2 -1; 0 -1 2] ## Short answer Using power method with initial vector [1,1,1], after iterations the dominant eigenvalue converges to 3.4142 and corresponding eigenvector to [-0.7071, 1.0000, -0.7071]. ## Answer **Power method:** to find the dominant (largest in magnitude) eigenvalue λ and its eigenvector x of A, start with a trial vector x(0), compute A·x(k) repeatedly and scale it by its largest component. The scale factor tends to λ and the scaled vector tends to the eigenvector. #### Step 1: Matrix and starting vector ```text A = 2 -1 0 -1 2 -1 0 -1 2 ``` Take x(0) = [1, 1, 1]ᵀ. Iterate: A·x(k) = λ(k)·x(k+1), where λ(k) is the component of largest magnitude in A·x(k). #### Step 2: Iterations | k | x(k) | A·x(k) | λ(k) (largest component) | x(k+1) = A·x(k)/λ(k) | |---|---|---|---|---| | 1 | [1.0000, 1.0000, 1.0000] | [1.0000, 0.0000, 1.0000] | 1.0000 | [1.0000, 0.0000, 1.0000] | | 2 | [1.0000, 0.0000, 1.0000] | [2.0000, -2.0000, 2.0000] | 2.0000 | [1.0000, -1.0000, 1.0000] | | 3 | [1.0000, -1.0000, 1.0000] | [3.0000, -4.0000, 3.0000] | -4.0000 | [-0.7500, 1.0000, -0.7500] | | 4 | [-0.7500, 1.0000, -0.7500] | [-2.5000, 3.5000, -2.5000] | 3.5000 | [-0.7143, 1.0000, -0.7143] | | 5 | [-0.7143, 1.0000, -0.7143] | [-2.4286, 3.4286, -2.4286] | 3.4286 | [-0.7083, 1.0000, -0.7083] | | 6 | [-0.7083, 1.0000, -0.7083] | [-2.4167, 3.4167, -2.4167] | 3.4167 | [-0.7073, 1.0000, -0.7073] | | 7 | [-0.7073, 1.0000, -0.7073] | [-2.4146, 3.4146, -2.4146] | 3.4146 | [-0.7071, 1.0000, -0.7071] | | 8 | [-0.7071, 1.0000, -0.7071] | [-2.4143, 3.4143, -2.4143] | 3.4143 | [-0.7071, 1.0000, -0.7071] | | 9 | [-0.7071, 1.0000, -0.7071] | [-2.4142, 3.4142, -2.4142] | 3.4142 | [-0.7071, 1.0000, -0.7071] | | 10 | [-0.7071, 1.0000, -0.7071] | [-2.4142, 3.4142, -2.4142] | 3.4142 | [-0.7071, 1.0000, -0.7071] | From the last two rows, λ and the vector are unchanged to 4 decimal places. **Answer:** dominant eigenvalue λ ≈ 3.4142; corresponding eigenvector ≈ [-0.7071, 1.0000, -0.7071]ᵀ (any multiple of it is also an eigenvector). ## Check yourself The answer's key points as multiple-choice questions, with the correct option and why. 1. Starting vector used: A. [1, 1, 1] B. [0,0,0] C. [1,0,0] D. [2,-1,0] Answer: A. Non-zero trial vector. (Key point: Start) 2. A*[1,1,1] equals: A. [3,3,3] B. [1, 0, 1] C. [2,2,2] D. [1,1,1] Answer: B. Row sums: 2-1, -1+2-1, -1+2. (Key point: First product) 3. Dominant eigenvalue is about: A. 2.0000 B. 4.0000 C. 3.4142 D. 1.0000 Answer: C. Equals 2 + sqrt(2). (Key point: Eigenvalue) 4. The eigenvector is: A. [1,1,1] B. [1,0,1] C. [0.7071,0,0.7071] D. [-0.7071, 1, -0.7071] Answer: D. Any multiple is valid. (Key point: Eigenvector) ## Related in Numerical Method - [Power method pseudo-code](https://www.bctengineer.com/question.md?key=semester_4%2Fnumerical_method%3Apower-method-algorithm) — Numerical Method, chapter 3, asked 2× - [Power method: 1,4,-1 matrix](https://www.bctengineer.com/question.md?key=semester_4%2Fnumerical_method%3Apower-method-1-4-minus1) — Numerical Method, chapter 3, asked 3× - [Power method: 15,-4,-3 matrix](https://www.bctengineer.com/question.md?key=semester_4%2Fnumerical_method%3Apower-method-15-4-3) — Numerical Method, chapter 3, asked 2× - [Power method: 2,5,1 matrix](https://www.bctengineer.com/question.md?key=semester_4%2Fnumerical_method%3Apower-method-2-5-1) — Numerical Method, chapter 3, asked 4× - [Power method: 1,2,1 matrix](https://www.bctengineer.com/question.md?key=semester_4%2Fnumerical_method%3Apower-method-1-2-1) — Numerical Method, chapter 3, asked 2× - [Power method: 25,1,2 matrix](https://www.bctengineer.com/question.md?key=semester_4%2Fnumerical_method%3Apower-method-25-1-2) — Numerical Method, chapter 3, asked 2× ## Related in other subjects - [Tridiagonal matrix, river stage and water table](https://www.bctengineer.com/question.md?key=semester_8%2Fcomputational_techniques_in_civil_engineering%3Atridiagonal-matrix-river-aquifer) — Computational Techniques in Civil Engineering, chapter 7, asked 2×