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# ch.2 - Chapter 2 Eigenvalues Eigenvectors Diagonalization...

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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 = 0 . 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 0 is an eigenvalue of A , then the eigenvectors of A corresponding to λ 0 are exactly those NONTRIVIAL solutions of the system ( A - λ 0 I n ) x = 0 . 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 ! = 0 0 ! . Calculation shows that the nontrivial solution to this system is given by x y ! = α 1 8 ! for any α R \ { 0 } . 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 ! · x y ! = 0 0 ! , which gives the nontrivial solution x y ! = β 1 2 ! for any β R \ { 0 } , and so we have obtained eigenvectors of A corresponding to the eigenvalue λ 2 = 4 . 2

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Example 2.4 Let A = 5 - 4 1 1 ! . Find all the eigenvalues of A and their corresponding eigenvectors.
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ch.2 - Chapter 2 Eigenvalues Eigenvectors Diagonalization...

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