Inverse-power-method.pdf
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1) The document discusses using the inverse power method to find the smallest eigenvalue of a matrix. It shows that if λ is an eigenvalue of an invertible matrix A, then 1/λ is an eigenvalue of A^-1. 2) As an example, it considers the matrix B and shows how to derive its two eigenvalues using its characteristic equation. 3) It then applies the inverse power method to the matrix B^-1 to iteratively calculate the eigenvector corresponding to the smallest eigenvalue of B, converging on the value of 0.225.
Inverse-power-method.pdf
- 1. Isaac Amornortey Yowetu Inverse Power Method May 5, 2022
- 2. Theorem (Inverse Eigenvalues) Suppose that λ, x is an eigenpair of A. If λ 6= 0, then 1 λ , x is an eigenpair of the matrix A−1 Suppose A is an invertible matrix or non-singular matrix. Ax = λx (1) A−1 Ax λ = λA−1 x λ (2) 1 λ x = A−1 x (3) 2 / 9
- 3. Finding least Eigenvalue by the Inverse Power M Considering matrices A and B A = −1 3 2 2 3 −1 ! Det(A)=0. Hence, it a singular or non-invertible matrix. B = −1 3 2 1 −1 ! Det(B) = −1 2 . Hence, it a non-singular or invertible matrix. 3 / 9
- 4. Problem Find the smallest eigenvalue of the matrix using the inverse power method B = −1 3 2 1 −1 ! Take the initail approximation x0 = (1, 1), 4 / 9
- 5. Finding Eigenvalues Using a Characteristic Equ B = −1 3 2 1 −1 ! Using the characteristic equation: ρ(A) = λ2 + 2λ − 0.5 = 0 λ = −b ± q b2 − 4(ac) 2a λ = −2 ± q 4 − 4(−0.5) 2 λ1 = −2.2247, λ2 = 0.2247 5 / 9
- 6. Solution A = B−1 (4) = 2 3 2 2 ! (5) First Iteration yields Ax0 = 2 3 2 2 ! 1 1 ! = 5 4 ! (6) By applying scaling we have the first approx. x1 = 1 5 5 4 ! = 1.00 0.80 ! (7) 6 / 9
- 7. Solution Second Iteration yields Ax1 = 2 3 2 2 ! 1 0.80 ! = 4.4 3.6 ! (8) By applying scaling we have the second approx. x2 = 1 4.4 4.4 3.6 ! = 1.00 0.82 ! (9) 7 / 9
- 8. Solution Third Iteration yields Ax2 = 2 3 2 2 ! 1.00 0.82 ! = 4.46 3.64 ! (10) By applying scaling we have the third approx. x3 = 1 4.46 4.46 3.64 ! = 1.00 0.82 ! (11) 8 / 9
- 9. Solution Fourth Iteration yields Ax3 = 2 3 2 2 ! 1.00 0.82 ! = 4.46 3.64 ! (12) By applying scaling we have the fourth approx. x4 = 1 4.46 4.46 3.64 ! = 1.00 0.82 ! (13) The smallest eigenvalue is: x4 = 1 4.46 = 0.225 9 / 9