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Unformatted text preview: Chapter 2. Eigenvalues, Eigenvectors, Diagonalization 2.1. Eigenvalues and Eigenvectors Definition 2.1 Let A ∈ M n ( R ) . A nonzero vector x ∈ M n × 1 ( R ) is called an eigenvector of A if A x is a scalar multiple of x , that is, A x = λ x (2.1) for some λ ∈ R . The scalar λ is called an eigenvalue of A and x is said to be an eigenvector of A corresponding to λ . In other words, the effect of multiplying x by the matrix A is equivalent to multiplying x by the scalar λ . The geometric interpretation is that an eigenvector of a given matrix is a vector which is rescaled by a constant (eigenvalue), possibly reverse the direction, when we apply the matrix to the vector. For example, an eigenvalue 2 means that the eigenvector is doubled in length, while an eigenvalue 1 means that the eigenvector stays the same in length but is reversed in direction. Example 2.2 Let A = 6 1 16 4 ! and x = 1 8 ! . Then A x = 6 1 16 4 ! · 1 8 ! = 2 16 ! = 2 1 8 ! = 2 x . Hence 2 is an eigenvalue of A and the vector 1 8 ! is an eigenvector of A corresponding to the eigenvalue 2 . 2 Remark Note that if x is an eigenvector of A corresponding to λ , then k x , k 6 = 0 , is also an eigenvector of A corresponding to λ . Equation (2.1) can be written as ( A λI n ) x = . Let B = A λI n . If λ is an eigenvalue of A , then there exists a nonzero vector x such that B x = 0 . Proposition 1.46 then implies that B is not invertible, that is, B is singular, or det( B ) = 0 . Therefore, in order to find eigenvalues of A , we need to solve det( A λI n ) = 0 . (2.2) This equation is called the characteristic equation of A . The following gives the procedures of finding eigenvalues and their corresponding eigenvectors. Procedures to find eigenvalues and their corresponding eigenvectors 1. Find all the eigenvalues of A by solving the characteristic equation (2.2). 2. If λ is an eigenvalue of A , then the eigenvectors of A corresponding to λ are exactly those NONTRIVIAL solutions of the system ( A λ I n ) x = . Example 2.3 Let A = 6 1 16 4 ! . Find all the eigenvalues of A and their corresponding eigenvectors. Solution. The characteristic equation of A is given by 6 λ 1 16 4 λ = 0 . Hence the eigenvalues of A are the roots of the equation λ 2 2 λ 8 = 0 , that is, λ 1 = 2 and λ 2 = 4 . For λ 1 = 2 , we have A + 2 I 2 = 8 1 16 2 ! . To find eigenvectors corresponding to λ 1 = 2 , we solve 8 1 16 2 ! · x y ! = ! . Calculation shows that the nontrivial solution to this system is given by x y ! = α 1 8 ! for any α ∈ R \{ } . Thus every vector of the form α 1 8 ! is an eigenvector of A corresponding to the eigenvalue λ 1 = 2 . Similarly, for λ 2 = 4 , we solve 2 1 16 8 !...
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This note was uploaded on 12/20/2011 for the course MATH 1813 taught by Professor Wu during the Fall '11 term at HKU.
 Fall '11
 Wu
 Math, Eigenvectors, Vectors

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