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Lecture 33.pdf

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MATH 136
Robert Sproule

Friday, March 28 − Lecture 33 : Eigenvectors associated to an eigenvalue. Concepts: 1. Define the family of all eigenvectors associated to an eigenvalue. 2. Find all eigenvectors associated to an eigenvalue. 3. Define an eigenspace. 33.1 Theorem − Let λ be a number. The number λ is an eigenvalue ofA if and only if 1 1 the matrix equation Ax = λ x w1th variable x has a non-trivial solution. Proof : λ 1s an eigenvalue ofA ⇔ λ is a 1olution to det(A − λI) = 0 ⇔ det(A − λ I)1= 0 ⇔ A − λ I 1s not invertible ⇔ (A − λ I)1 = 0 has a non-trivial solution ⇔ Ax − λ Ix1= 0 has a non-trivial solution ⇔ Ax = λ Ix 1as a non-trivial solution ⇔ Ax = λ x h1s a non-trivial solution. Different ways of viewing an eigenvalue. The following are equivalent: 1. λ i1 an eigenvalue ofA 2. (A − λ I1x = 0 has a non-trivial solution 3. Ax = λ x 1as a non-trivial solution 4. Null(A − λ I)1≠ {0} has a non-trivial solution 33.1.1 Example − Show that 1 is an eigenvalue of the following matrix A Solution: By the above theorem it suffices to verify that the following homogeneous system(A – (1)I)x = 0. has a non-trivial solution. This coefficient matrix easily row-reduces to Since (A – (1)I)x = 0 has a non-trivial solution then 1 is an eigenvalue of A. Then Ax = x has a non-trivial solution. Similarly, Ax = 2x and Ax = 3x both have non-trivial solutions and so 2 and 3 are eigenvalues of A. 33.1.2 Different ways of viewing an eigenvalue. We have the following different ways of viewing an eigenvalue of A: λ1= eigenvalue of A ⇔ det(A − λ I) = 0 1 ⇔ A − λ I 1s not invertible, ⇔ (A − λ I)1 = 0 has a non-trivial solution. ⇔ Null(A − λ I) 1ontains infinitely many vectors ⇔ Ax = λ x for infinitely many vectors x 1 33.2. Definition − Suppose λ is a1 eigenvalue of n × n matrix A. Then any non-zero vector x in Null(A − λ I) is called an eigenvectorof A corresponding to λ . 1 1 1 33.2.1 Remark – By definition of “eigenvector of a matrix A” an eigenvector always exists in relation to some eigenvalue. So to say that x is an eigenvector of the matrix 1 A invites the following question: “x is 1ssociated to which of the eigenvalues of A?”. An eigenvector of A can only correspond to a single eigenvalue. If x is an 1 eigenvector of A associated to the eigenvalues λ and λ1of A th2n λ = λ . Thi1 is 2 because x c1nnot be the zero vector. To see this consider: The expression “λ 1 1= λ 2 1mplies λ = λ1” si2ce an eigenvector cannot be the zero vector. 33.2.2 Remark – Extending the notion of “eigenvalue” so that it applies to linear mappings. The notions of “eigenvalue” and “eigenvector” of a matrix A can be extended to a linear map (transformation) T : ℝ → ℝ .n n Definition –We say that λ is 1n eigenvalue of the linear map T: ℝ → ℝ if n n T(x) = λ x1 has a non-trivial solution for x. We say thatx 1 is an eigenvector of the linear transformation T corresponding to the eigenvalue λ 1 if x1is a non-trivial solution to T(x) = λ x1 n n - If λ i1 an eigenvalue of T: ℝ → ℝ and x is an eigenvec1or corresponding to λ 1hen, “T maps x to a1scalar multiple λ x of x ”1 1 1 2 2 Example – Suppose R : ℝ → ℝ πs the linear transformation which rotates points in ℝ counterclockwise about the origin by an angle of π radians. Suppose we seek an eigenvalue and its associated eigenvectors for this linear map. See that if x is any ordered pair in ℝ , R (x) = −x. Then λ = −1 is an eigenvalue of π 1 2 R π The eigenvectors associated to λ = −1 ar1 all non-zero vectors in ℝ . The set of all eigenvectors corresponding to an eigenvalue – If λ is an eigen1alue of a matrix A, all the eigenvectors associated to λ form a1set or “package”. This set is the nullspace, Null(A − λ I ), of the mat
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