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§4.4 Similarity and Diagonalization

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§4.4 Similarity and Diagonalization An illustrative example Similar matrices Diagonalization §4.4 Similarity and Diagonalization Tom Lewis Fall Semester 2020 An illustrative example Similar matrices Diagonalization Outline An illustrative example Similar matrices Diagonalization An illustrative example Similar matrices Diagonalization An illustrative example An illustrative example Similar matrices Diagonalization Problem Let T be the transformation in the plane that reflects a vector through the line passing through the origin with direction [3, 4]. Find the matrix of the transformation T. An illustrative example Similar matrices Diagonalization A “natural” basis for T T(v) u=[3,4] v=[4,-3] An illustrative example Similar matrices Diagonalization In summary Let -0.28 0.96 3 4 1 0 A = , P = , D = 0.96 0.28 4 -3 0 -1 Then A = PDP -1 or P -1AP = D A and D represent the same transformation, T, but with respect to different bases. An illustrative example Similar matrices Diagonalization Similar matrices An illustrative example Similar matrices Diagonalization The similarity condition above can also be expressed as A = PBP -1. A and B are the same transformation represented through different bases. An illustrative example Similar matrices Diagonalization Example Let 10 12 1 0 A = and B = -6 -7 0 2 Observe that 10 12 4 -3 1 0 -2 -3 = . -6 -7 -3 2 0 2 -3 -4 This shows that A ∼ B. An illustrative example Similar matrices Diagonalization Properties (a), (b), and (c) are the defining properties of an equivalence relation: matrix similarity is an example of an equivalence relation. An illustrative example Similar matrices Diagonalization An illustrative example Similar matrices Diagonalization Example The matrices 1 2 1 -1 A = and B = 0 1 1 2 are not similar: det A = 1 and det B = 3 An illustrative example Similar matrices Diagonalization Diagonalization An illustrative example Similar matrices Diagonalization Equivalently, A = PDP -1. An illustrative example Similar matrices Diagonalization Problem Let D be an n × n diagonal matrix and let P be an invertible n × n matrix with columns p1,..., pn . Thus 2 3 λ1 ··· 0 6 . .. 7 D = 4 . 0 5 and P = p1 ··· pn . 0 ··· λn Let A = PDP -1. For each i, 1 6 i 6 n, show that Api = λi pi . Thus A has n linearly independent eigenvectors. An illustrative example Similar matrices Diagonalization Example Suppose 3 4 -4 0 3 -4 A = . 2 3 0 7 -2 3 Then A has eigenvalues -4 and 7 and corresponding eigenvectors 3 4 and . 2 3 An illustrative example Similar matrices Diagonalization 1. If A = PDP -1, then A has n linearly independent eigenvectors. 2. If A has n linearly independent eigenvectors, then A = PDP -1 for some invertible matrix P and some diagonal matrix D. An illustrative example Similar matrices Diagonalization Problem -13 24 Show that the matrix A = is diagonalizable. -8 15 An illustrative example Similar matrices Diagonalization Solution Observe that -13 - λ 24 2 pA(λ) = = λ - 2λ - 3 = (λ + 1)(λ - 3) -8 15 - λ The spectrum of A is f-1, 3g. Let p1 and p2 be the eigenvectors corresponding to λ = -1 and λ = 3. The eigenvectors p1 and p2 are linearly independent since they correspond to distinct eigenvalues; thus, A is diagonalizable. An illustrative example Similar matrices Diagonalization An illustrative example Similar matrices Diagonalization Problem Show that the matrix 2 5 6 0 3 A = 4-3 -4 0 5 . -3 -3 -1 is diagonalizable. An illustrative example Similar matrices Diagonalization Solution The characteristic polynomial of A is 2 pA(λ) = -(λ + 1) (λ - 2). Thus A has only two distinct eigenvalues: -1 and 2. 2-13 203 2-23 E-1 = span 4 1 5 , 405 and E2 = span 4 1 5 8 9 8 9 < 0 1 = < 1 = : ;λ = -1 λ = 2 : ; alg. mult. 2 1 geo. mult. 2 1 An illustrative example Similar matrices Diagonalization The final form Here is the diagonal representation of the matrix A: 2 5 6 0 3 2-1 0 -23 2-1 0 03 2 1 2 03 4-3 -4 0 5 = 4 1 0 1 5 4 0 -1 05 4 1 1 15 -3 -3 -1 0 1 1 0 0 2 -1 -1 0 An illustrative example Similar matrices Diagonalization An illustrative example Similar matrices Diagonalization Problem 3 1 Let A = . Show that A is not diagonalizable. 0 3 An illustrative example Similar matrices Diagonalization Solution The characteristic polynomial of A is 2 pA(λ) = (λ - 3) . Thus A has only one eigenvalue: 3. 1 E = span 3 0 λ = 3 alg. mult. 2 geo. mult. 1 The matrix A is not diagonalizable. An illustrative example Similar matrices Diagonalization Problem 4 -6 The matrix A = has has eigenvalues λ = 2 and λ = 1 1 -1 3 2 with corresponding eigenvectors and . 1 1 1. Find a diagonalization of A. 2. Express A5. 3. Find the limit of 2-n An as n ! . 1.
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