Fall 2001 16.3110-3 Eigenvalues and eigenvectors Recall that the eigenvalues of A are the same as the roots of the characteristic equation (page A is an eigenvalue of A if det(AI-A)=0 which is true iff there exists a nonzero v(eigenvector) for which Au=入 Repeat the process to find all of the eigenvectors. Assuming that the n eigenvectors are linearly independent A AT=TA TTAT=A . One word of caution: Not all matrices are diagonalizable 01 det(sI-A)= only one eigenvalue s=0 (repeated twice). The eigenvectors solve 「01 0 00 eigenvectors are 71 of the form0/,n≠0→ would only be one Need the Jordan Normal Form to handle this case(section 3.7.3)Fall 2001 16.31 10–3 Eigenvalues and Eigenvectors • Recall that the eigenvalues of A are the same as the roots of the characteristic equation (page 9–1) • λ is an eigenvalue of A if det(λI − A)=0 which is true iff there exists a nonzero v (eigenvector) for which (λI − A)v = 0 ⇒ Av = λv • Repeat the process to find all of the eigenvectors. Assuming that the n eigenvectors are linearly independent Avi = λivi i = 1,...,n A  v1 ··· vn  =  v1 ··· vn    λ1 ... λn   AT = TΛ ⇒ T −1 AT = Λ • One word of caution: Not all matrices are diagonalizable A = 0 1 0 0 det(sI − A) = s2 only one eigenvalue s = 0 (repeated twice). The eigenvectors solve 0 1 0 0 r1 r2 = 0 eigenvectors are of the form r1 0 , r1 = 0 → would only be one. • Need the Jordan Normal Form to handle this case (section 3.7.3)