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EIGENVECTOR CALCULATION Let A have an approximate eigenvalue λ, so that A − λI is almost singular. How do we find a corresponding eigenvector? If the eigenvalue is of multiplicity 1, then in linear algebra courses we usually just try to solve the linear system (A − λI ) x = 0 Oversimplifying, we usually drop one of the equations, arbitrarily assign one of the components of x, and then solve for the remaining components. In numerical computations, this almost always leads to at least one of the eigenvectors being obtain inaccurately due to solving an ill-conditioned linear system. Consequently, another approach is usually used to find the eigenvector x. INVERSE ITERATION Let λ be an approximate eigenvalue of A, corresponding to some true eigenvalue λk for A. Let xk denote the associated eigenvector. Choose an initial estimate z (0) ≈ xk , often using a random number generator. The inverse iteration method is defined by (A − λI ) w(m+1) = z (m) (m+1) w z (m+1) = (m+1) w ∞ for m = 0, 1, 2, ... It is important that λ not be exactly a true eigenvalue, or else the matrix A − λI will be singular. EXAMPLE Let 6 4 3 A= 4 3 2 3 2 1 The eigenvalues are . λ1 = 0.23357813629678 . λ2 = −0.42027581011042 . λ3 = 10.18669767381363 The corresponding normalized eigenvectors are −0.64061130860597 1.00000000000000 , −0.10198831463966 0.37938985766816 0.14105311855297 −1.00000000000000 1.00000000000000 0.68921959524450 0.47660643094521 In all cases, with a random z (0), we had z (1) was the correct answer to the digits shown. CONVERGENCE Let A have a diagonal Jordan canonical form, P −1AP = D = diag [λ1, ..., λn ] with P = [x1, ..., xn] the matrix of eigenvectors (assumed to be normalized with xi∞ = 1). For the given z (0), expand it in the basis of eigenvectors {x1, ..., xn}: z (0) = n j=1 α j xj Let λ ≈ λk for some k; and for simplicity, assume λk is a simple eigenvalue of A. Moreover, assume αk = 0. This is generally assured by using a random number generator to define z (0). What are the eigenvalues and eigenvectors of A − λI ? (A − λI ) xi = (λi − λ) xi, i = 1, ..., n Also, this implies (A − λI )−1 xi = 1 xi , λi − λ i = 1, ..., n Our inverse iteration method (A − λI ) w(m+1) = z (m) w(m+1) (m+1) z = w(m+1) ∞ is a power method. Simply write it in the form w(m+1) = (A − λI )−1 z (m) In §9.2 on the power method, replace the role of A by (A − λI )−1. Then from (9.2.4) of that section, z (m) = σm (A − λI )−m z (0) , −m (0) z (A − λI ) ∞ m≥0 with |σm| = 1. Using this formula and the earlier (A − λI )−1 xi = we have 1 xi , λi − λ n 1 λj − λ j=1 (A − λI )−m z (0) = = 1 λk − λ m i = 1, ..., n m α j xj m n λk − λ α k xk + j=1 j=k λj − λ α j xj Then α k xk + z (m) = σm m n λk − λ j=1 j=k λj − λ α j xj m n λ − λ k α x + α x j j k k j=1 λj − λ j=k ∞ If |λk − λ| λj − λ , j= k then z (m) ≈ σmxk with only a small value of m. The error with iteration decreases by a factor of λ − λ max k j=k λj − λ This explains the rapid convergence of our numerical example.