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# set6 - \$ Set 6 Symmetric Eigenvalue Problem Part 2 Kyle A...

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a39 a38 a36 a37 Set 6: Symmetric Eigenvalue Problem Part 2 Kyle A. Gallivan Department of Mathematics Florida State University Numerical Linear Algebra 1 Fall 2010 1

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a39 a38 a36 a37 Hermitian Tridiagonal Eigenvalue Problem The Hermitian (symmetric) eigenproblem has a rich theory behind it and there are many ways to approximate it. We will concentrate on the so-called symmetric QR algorithm for real symmetric matrices. The discussion comprises two parts: the explicit QR and the implicit QR algorithms. We start by assuming a simpler symmetric problem, i.e., the symmetric tridiagonal eigenvalue problem. 2
a39 a38 a36 a37 Symmetric Tridiagonal Eigenvalue Problem Definition 6.1. T R n × n is a symmetric tridiagonal matrix if and only if T T = T and e T i Te j = 0 when i > j + 1 . In other words, e T i Te j negationslash = 0 only when i = j (the main diagonal), i = j + 1 (the subdiagonal), or i + 1 = j (the superdiagonal). Example for n = 5 : T = 10 5 0 0 0 5 52 23 0 0 0 23 33 4 0 0 0 4 104 12 0 0 0 12 85 3

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a39 a38 a36 a37 Transformation to Tridiagonal Problem To handle a symmetric matrix M we add a preprocessing step that creates an orthogonal similarity transformation, H . H transforms M to a symmetric tridiagonal matrix T . We solve the symmetric tridiagonal EVP to find T = Q Λ Q T . This yields H T MH = T T = Q Λ Q T M = HQ Λ Q T H T = U Λ U T 4
a39 a38 a36 a37 Householder Reflector Theorem 6.1. Given a vector v R n , and γ = ±bardbl v bardbl 2 , if x = v + γe 1 and α = 2 bardbl x bardbl 2 2 . then Hv = ( I + αxx T ) v = γe 1 The sign of γ is chosen as sgn ( e T 1 v ) for numerical reasons. 5

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a39 a38 a36 a37 Example v = 1 1 1 1 , γ = 2 , x = 3 1 1 1 , bardbl x bardbl 2 2 = 12 , α = 1 6 H = 1 6 3 3 3 3 3 5 1 1 3 1 5 1 3 1 1 5 Hv = 2 e 1 6
a39 a38 a36 a37 Householder Reflector Easy to generalize to introducing 0 ’s in positions i + 1 , · · · , n leaving elements 1 , · · · , i 1 untouched and element i an updated nonzero value: H i v = I i 1 0 0 ˜ H v 1 v 2 = v 1 e 1 γ ˜ H = I n i +1 + αxx T , x R n i +1 H i = I i 1 0 0 I n i +1 + α 0 x parenleftBig 0 x T parenrightBig 7

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a39 a38 a36 a37 Reduction to Tridiagonal Form Use pre- and postmultiplication by Householder reflectors. T = H T MH = H T n 2 H T n 2 · · · H T 1 MH 1 · · · H n 2 gives similar structure to the first row. 8
a39 a38 a36 a37 Reduction to Tridiagonal Form Choose H 1 as the reflector that introduces 0 ’s into positions 3 through n in Me 1 M = H T 1 M =

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• Fall '06
• gallivan
• Orthogonal matrix, symmetric tridiagonal matrix, Tridiagonal, Tridiagonal Problem, symmetric tridiagonal eigenvalue, tridiagonal eigenvalue problem

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set6 - \$ Set 6 Symmetric Eigenvalue Problem Part 2 Kyle A...

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