Chapter 6 Eigenvalues and Eigenvectors
Chapter 6 Eigenvalues and Eigenvectors Po-Ning Chen, Professor Department of Electrical and Computer Engineering National Chiao Tung University Hsin Chu, Taiwan 30010, R.O.C. 6.1 Introduction to eigenvalues 6-1 Motivations • The static system problem of Ax = b has now been solved, e.g., by Gauss- Jordan method or Cramer’s rule. • However, a dynamic system problem such as Ax = λx cannot be solved by the static system method. • To solve the dynamic system problem, we need to find the static feature of A that is “unchanged” with the mapping A. In other words, Ax maps to itself with possibly some stretching (λ>1), shrinking (0 <λ<1), or being reversed (λ<0). • These invariant characteristics of A are the eigenvalues and eigenvectors. Ax maps a vector x to its column space C(A). We are looking for a v ∈ C(A) such that Av aligns with v. The collection of all such vectors is the set of eigen- vectors. 6.1 Eigenvalues and eigenvectors 6-2 Conception (Eigenvalues and eigenvectors): The eigenvalue-eigenvector pair (λi, vi) of a square matrix A satisfies Avi = λivi (or equivalently (A − λiI)vi = 0) where vi = 0 (but λi can be zero), where v1, v2,..., are linearly independent. How to derive eigenvalues and eigenvectors? • For 3 × 3 matrix A, we can obtain 3 eigenvalues λ1,λ2,λ3 by solving 3 2 det(A − λI)=−λ + c2λ + c1λ + c0 =0. • If all λ1,λ2,λ3 are unequal, then we continue to derive: Av1 = λ1v1 (A − λ1I)v1 = 0 v1 =theonly basis of the nullspace of (A − λ1I) Av λ v ⇔ A − λ I v ⇔ v A − λ I 2 = 2 2 ( 2 ) 2 = 0 2 =theonly basis of the nullspace of ( 2 ) Av3 = λ3v3 (A − λ3I)v3 = 0 v3 =theonly basis of the nullspace of (A − λ3I) and the resulting v1, v2 and v3 are linearly independent to each other.
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