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Unformatted text preview: Lecture Note 111: Nov 21  Nov 24, 2006 Dr. Jeff ChakFu WONG Department of Mathematics Chinese University of Hong Kong jwong@math.cuhk.edu.hk MAT 2310 Linear Algebra and Its Applications Fall, 2006 Produced by Jeff ChakFu WONG 1 E IGNEVALUES , E IGENVECTORS , AND D IAGONALIZATION 1. Eignevalues and Eigenvectors 2. Diagonalization 3. Diagonalization of Symmetric Matrices EIGNEVALUES, EIGENVECTORS, AND DIAGONALIZATION 2 D IAGONALIZATION OF S YMMETRIC M ATRICES DIAGONALIZATION OF SYMMETRIC MATRICES 3 In this part we consider the diagonalization of symmetric matrices (an n × n matrix A with real entries that A = A T ). We restrict our attention to this case because symmetric matrices arise in many applied problems. DIAGONALIZATION OF SYMMETRIC MATRICES 4 As an example of such a problem, consider the task of identifying the conic represented by the equation 2 x 2 + 2 xy + y 2 = 9 , which can be written in matrix form as h x y i 2 1 1 2 x y = 9 Observe that the matrix used here is a symmetric matrix. We shall merely remark here that the solution calls for the determination of the eigenvalues and eigenvectors of the matrix 2 1 1 2 . The x and yaxes are then rotated to a new set of axes, which lie along the eigenvectors of the matrix. In the new set of axes, the given conic can be identified readily. DIAGONALIZATION OF SYMMETRIC MATRICES 5 Theorem 0.1 All the roots of the characteristic polynomial of a symmetric matrix are real numbers. We omit the proof of the following important theorem (see D. R. Hill, Experiments in Computational Matrix Algebra, New York: Random House, 1988) DIAGONALIZATION OF SYMMETRIC MATRICES 6 Theorem 0.2 If A is a symmetric matrix, then eigenvectors that are associated with distinct eigenvalues of A are orthogonal. Proof: First, we shall let the reader verify the property that if x and y are vectors in R n , then ( A x ) · y = x · ( A T y ) . Now let x 1 and x 2 be eigenvectors of A associated with the distinct eigenvalues λ 1 and λ 2 of A . We then have A x 1 = λ 1 x 1 , A x 2 = λ 2 x 2 . Now λ 1 ( x 1 · x 2 ) = ( λ 1 x 1 ) · x 2 = ( A x 1 ) · x 2 = x 1 · ( A T x 2 ) = x 1 · ( A x 2 ) = x · ( λ 2 x 2 ) = λ 2 ( x 1 · x 2 ) where we have used the fact that A = A T . DIAGONALIZATION OF SYMMETRIC MATRICES 7 Thus λ 1 ( x 1 · x 2 ) = λ 2 ( x 1 · x 2 ) and subtracting, we obtain 0 = λ 1 ( x 1 · x 2 ) λ 2 ( x 1 · x 2 ) = ( λ 1 λ 2 )  {z } 6 =0 ( x 1 · x 2 )  {z } =0 . Since λ 1 6 = λ 2 , we conclude that x 1 · x 2 = 0 , so x 1 and x 2 are orthogonal. DIAGONALIZATION OF SYMMETRIC MATRICES 8 Example 1 Given the symmetric matrix A =  2 2 2 3 , we find that the characteristic polynomial of A is (verify) f ( λ ) = ( λ + 2)( λ 4)( λ + 1) , so the eigenvalues of A are λ 1 = 2 , λ 2 = 4 , λ 3 = 1 ....
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This note was uploaded on 10/15/2009 for the course MATHEMATIC MAT2310B taught by Professor Jeffwong during the Spring '09 term at CUHK.
 Spring '09
 JeffWong
 Linear Algebra, Algebra

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