ERG2013_lecture_24

Matrices have the characteristic polynomial

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Unformatted text preview: implies diagonalizable 35 kshum The converse is false If a matrix A is similar to a diagonal matrix, then A may have repeated eigenvalues. Eigenvalue = 2, 2,–3. Diagonalizable matrix may have repeated eigenvalues 36 kshum Facts Distinct eigenvalues Eigenvalues not all distinct 37 diagonalizable may or may not be diagonalizable kshum Calculation of inverse required in diagonalization May take a lot of time to compute when the size of matrix is large 38 kshum Symmetric matrix Definition: A matrix M is symmetric if MT = M. Examples: Line of symmetry 39 kshum Diagonalization of symmetric matrix Transpose instead of inverse 40 kshum Orthogonal matrix A matrix M is called orthogonal if Each column has norm 1 MT I M Dot product = 1 41 kshum Orthogonal matrix A matrix M is called orthogonal if Any two distinct columns are orthogonal Dot product = 0 42 kshum Orthogonal matrix If then M is also orthogonal Each row has norm 1 Any two distinct rows are orthogonal M Dot product = 0 43 MT kshum Theorem Two eigenvectors of a symmetric matrix are orthogonal if the corresponding eigenvalues are different. Let A be a symmetric matrix, AT = A. Let u is an eigenvector with eigenvalue λ, A u = λ u Let v is an eigenvector with eigenvalue µ. A v = µ v Suppose λ not equal to µ. (a scalar) 44 kshum Corollary 1. 2. 3. For an n× n symmetric A with n distinct eigenvalues, we can find n orthogonal eigenvectors. Form a matrix P with the n eigenvectors as columns. Normalize each column so that the norm of each column is equal to one. Then PT P = I and PTAP is a diagonal matrix, whose diagonal entries consist of the eigenvalues of A. Example: PT 45 A P kshum Principal axes theorem: all symmetric matrices are orthogonally diagonalizable A stronger statement is possible 1. For an n× n symmetric A with n distinct eigenvalues, we can find n orthogonal eigenvectors. 2. Form a matrix P with the n eigenvectors as columns. Normalize each column so that the norm of each column is equal to one. Then PT P = I and PTAP is a diagonal matrix, whose diagonal entries consist of the eigenvalues of A. 3. The conclusion in the previous page holds even without the hypothesis of distinct eigenvalues. 46 Proof omitted kshum Facts A has distinct eigenvalues Eigenvalues not all distinct Symmetric A, eigenvalues may or may not be distinct 47 P– 1AP = D may or may not be diagonalizable PTAP = D P TP = I kshum Application to conic sections Is x2 – 4xy +2y2 = 1 a ellipse, or a hyperbola? Rewrite using symmetric matrix Find the characteristic polynomial Solve for the eigenvalues 48 kshum Application to conic sections Diagonalize Change coordinates Hyperbola 49 kshum x2 – 4xy +2y2 = 1 15 10 y 5 0 -5 -10 -15 -15 -10 -5 0 5 10 x 50 kshum 15 2x2 + 2xy + y2 = 1 Rewrite using symmetric matrix Compute the characteristic polynomial Find the eigenvalues 51 kshum 2x2 + 2xy + y2 = 1 Diagonalize Columns of P are eigenvectors, normalized to norm 1. Change of variables 52 kshum 2x2 + 2xy + y2 = 1 v 0.8 0.6 0.4 y 0.2 0 -0.2 -0.4 -0.6 -0.8 -1 -0.5 0 0.5 1 u x 53 kshum Final 5 May, Wed UC Gym 9:30-11:30 Coverage: Kreyszig chapter 1.1~1.5, 2.1, 2.2, 2.4, 2.7~2.9, 4.0~4.5, 7.1~7.9, 8.1~8.5 Lecture notes, tutorial notes, assignments Close-book exam You may bring one single sheet of hand-written note 54 A4 size, double-sided kshum...
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This document was uploaded on 09/16/2013.

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