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# Lecture 34.pdf

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School
University of Waterloo
Department
Mathematics
Course
MATH 136
Professor
Robert Sproule
Semester
Winter

Description
Monday, March 31 − Lecture 34 : More on eigenvectors. Concepts: 1. Recognize that λ is a1 eigenvalue of Aif and only if 1/λ is an ei1envalue of A . -1 2. Define algebraic and geometric multiplicity of an eigenvalue. 34.1 Proposition − Let A be an invertible matrix. If λ is a1 eigenvalue of A then 1/λ is an 1 eigenvalue of A . Furthermore, x is an1eigenvector associated to 1/λ if and on1y if x is 1 an eigenvector of λ . 1ence, E 1/λ1= E λ1 Proof : Suppose matrix A is invertible. Then A does not have an eigenvalue equal to zero since (A – 0I)x = Ax = 0 only has the trivial solution (since A is invertible). λ 1is a eigenvalue of A. ⇔ the matrix equation Ax = λ x has1a non-trivial solution x 1 ⇔ Ax = 1 x f1 1some non-zero eigenvector −1 −1 ⇔ A Ax = A 1 x , 1 1 −1 ⇔ x =1A λ x . 1 1 ⇔ A x = (1/λ ) x . 1 1 1 ⇔ A x = (1/λ )x h1s a non-trivial solution ⇔ 1/λ i1 an eigenvalue ofA with x as an e1genvector. Hence, E 1/λ1= E λ1 34.2 Definitions Suppose λ is1an eigenvalue of the matrix A and E denotesλ1ts associated eigenspace. The dimension of E is λ1lled the geometric multiplicity of λ . 1 Recall that if λ is an eigenvalue of the matrix Aand det(A − λ I) = (λ + 1) = 0. we say 1 1 1 that the algebraic multiplicity of the eigenvalue λ 1 is k. Caution – One may suspect that the algebraic multiplicity of an eigenvalue λ must be 1 equal to its geometric multiplicity. But we will see that this is not always the case. 34.2.1 Example − Let A be the matrix, The determinant of A is 2 and so A is invertible. Compare the eigenvalues and eigenvectors A with the eigenvalues and eigenvectors of A . State the algebraic and geometric multiplicities of the eigenvalues for both matrices. Solution : To find the eigenvectors associated to 1 we solve for (A − 1I)x = 0: and get x = α(1, 0, 0) + β(0, 1, 0). Thus the basis of E ha1 two vectors. Since the dimension of E 1 is two, 1 is an eigenvalue of geometric multiplicity 2. We also easily determine in the same way that the basis for E is {(2,0,1)} . −1 We easily construct A and obtain: Proceeding as above we obtain λ 1,2= 1 and λ =3½, as predicted by the preceding theorem. We also obtain as the eigenspace associated to λ 1, 2= 1, E 1 α(1, 0, 0) + β(0, 1, 0) = span{(1, 0, 0) , (0, 1, 0)} and eigenspace associated to λ 1/2 = 1, E 1/2= span{(0,0,1)} Note that A and A have the same eigenspaces span{(1, 0, 0) , (0, 1, 0)} and span{(0,0,1)}. Remark – The reader may have noticed that the situation encountered when determining the basis of Null(A – λ I) isibit different from what we usually see when solving for systems Ax = b. So it is a good idea to practice this well. 34.2.2 Example − Verify that the matrix, has characteristic equation det(A − λI) = ( λ + 1) = 0 3 Hence the algebraic multiplicity of the eigenvalue λ = −1 i1 3. When the eigenvectors are calculated by solving (A − (−1)I)x = 0 we obtain as a solution set x = α(0, −3, 1) + β(1, 0, 0) . So E =1span{(0, −3, 1) , β(1, 0, 0)}. We see that the dimension of the eigenspace E is 1 2. It’s geometric multiplicity 2 while its algebraic multiplicity is 3. 34.2.4 Remarks − Suppose A is a 3 × 3 matrix. Then the matrix A may have − only one eigenvalue of algebraic multiplicity 3, − only 1 eigenvalue of algebraic multiplicity 1 and 1 of algebraic multiplicity 2 − three eigenvalues each of algebraic multiplicity 1. Furthermore, A may have − only one eigenspace
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