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Lecture_11_1

# Lecture_11_1 - Lecture Note 11-1 Nov 21 Dr Jeff Chak-Fu...

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Lecture Note 11-1: Nov 21 - Nov 24, 2006 Dr. Jeff Chak-Fu WONG Department of Mathematics Chinese University of Hong Kong [email protected] MAT 2310 Linear Algebra and Its Applications Fall, 2006 Produced by Jeff Chak-Fu WONG 1

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E IGNEVALUES , E IGENVECTORS , AND D IAGONALIZATION 1. Eignevalues and Eigenvectors 2. Diagonalization 3. Diagonalization of Symmetric Matrices E IGNEVALUES , E IGENVECTORS , AND D IAGONALIZATION 2
D IAGONALIZATION OF S YMMETRIC M ATRICES D IAGONALIZATION OF S YMMETRIC M ATRICES 3

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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. D IAGONALIZATION OF S YMMETRIC M ATRICES 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 y-axes 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. D IAGONALIZATION OF S YMMETRIC M ATRICES 5

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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) D IAGONALIZATION OF S YMMETRIC M ATRICES 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 . D IAGONALIZATION OF S YMMETRIC M ATRICES 7

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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. D IAGONALIZATION OF S YMMETRIC M ATRICES 8
Example 1 Given the symmetric matrix A = 0 0 - 2 0 - 2 0 - 2 0 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 . Then we can find the associated eigenvectors by solving the homogeneous system ( λ j I 3 - A ) x = 0 , j = 1 , 2 , 3 and obtain the respective eigenvectors (verify) x 1 = 0 1 0 , x 2 = - 1 0 2 , x 3 = 2 0 1 .

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