Numerical Methods Contents - SAM

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Random 8 × 8 circulant matrices C 1 , C 2 (→ Def. 7.1.3) Generated by MATLAB-command: C = gallery(’circul’,rand(n,1)); eigenvalues ✄ eigenvalue 5 4 3 2 1 0 −1 C 1 : real(ev) C 1 : imag(ev) C 2 : real(ev) C 2 : imag(ev) −2 0 1 2 3 4 5 6 7 8 9 index of eigenvalue Fig. 91 vector component value Circulant matrix 2, eigenvector 5 Circulant matrix 2, eigenvector 6 0.4 0.4 0.3 0.3 0.2 0.2 0.1 0.1 0 0 −0.1 −0.1 −0.2 −0.2 −0.3 −0.3 real part real part imaginary part imaginary part −0.4 −0.4 1 2 3 4 5 6 7 8 1 2 3 4 5 6 7 8 vector component index vector component index vector component value vector component value Circulant matrix 2, eigenvector 7 Circulant matrix 2, eigenvector 8 0.4 0.4 0.3 0.3 0.2 0.2 0.1 0.1 0 0 −0.1 −0.1 −0.2 −0.2 −0.3 −0.3 real part real part imaginary part imaginary part −0.4 −0.4 1 2 3 4 5 6 7 8 1 2 3 4 5 6 7 8 vector component index vector component index Observation: the different random circulant matrices have the same eigenvectors! Eigenvectors of C = gallery(’circul’,(1:128)’); : random 256x256 circulant matrix, eigenvector 2 random 256x256 circulant matrix, eigenvector 3 random 256x256 circulant matrix, eigenvector 5 random 256x256 circulant matrix, eigenvector 8 0.1 0.1 0.1 0.1 vector component value Little relationship between (complex!) eigenvalues can be observed, as can be expected from random matrices with entries ∈ [0, 1]. vector component value 0.08 0.06 0.04 0.02 0 −0.02 −0.04 vector component value 0.08 0.06 0.04 0.02 0 −0.02 −0.04 vector component value 0.08 0.06 0.04 0.02 0 −0.02 −0.04 vector component value 0.08 0.06 0.04 0.02 0 −0.02 −0.04 −0.06 −0.06 −0.06 −0.06 Eigenvectors of matrix C 1 : Ôº º¾ −0.08 real part imaginary part −0.1 0 20 40 60 80 100 120 140 vector component index −0.08 real part imaginary part −0.1 0 20 40 60 80 100 120 140 vector component index −0.08 real part imaginary part −0.1 0 20 40 60 80 100 120 140 vector component index −0.1 0 20 40 60 80 100 120 140 vector component index The eigenvectors remind us of sampled trigonometric functions cos(k/n), sin(k/n), k = 0,...,n−1! −0.08 real part imaginary part Ôº º¾ ✸ 0 −0.05 −0.1 Circulant matrix 1, eigenvector 1 0.4 0.3 0.2 Circulant matrix 1, eigenvector 2 0.4 0.3 0.2 Circulant matrix 1, eigenvector 3 0.4 0.3 0.2 Circulant matrix 1, eigenvector 4 Remark 7.2.3 (Why using K = C?). vector component value −0.15 −0.2 −0.25 vector component value 0.1 0 −0.1 vector component value 0.1 0 −0.1 vector component value 0.1 0 −0.1 Ex. 7.2.1: complex eigenvalues/eigenvectors for general circulant matrices. −0.3 −0.2 −0.2 −0.2 −0.35 −0.4 real part imaginary part 1 2 3 4 5 6 7 8 vector component index −0.3 −0.4 real part imaginary part 1 2 3 4 5 6 7 8 vector component index −0.3 −0.4 real part imaginary part 1 2 3 4 5 6 7 8 vector component index −0.3 −0.4 real part imaginary part 1 2 3 4 5 6 7 8 vector component index Recall from analysis: unified treatment of trigonometric functions via complex exponential function 0.4 Circulant matrix 1, eigenvector 5 0.4 Circulant matrix 1, eigenvector 6 0.4 Circulant matrix 1, eigenvector 7 0.4 Circulant matrix 1, eigenvector 8 exp(it) = cos(t) + i sin(t) , t ∈ R . vector component value 0.3 0.2 0.1 0 −0.1 vector component value 0.3 0.2 0.1 0 −0.1 vector component value 0.3 0.2 0.1 0 −0.1 vector component value 0.3 0.2 0.1 0 −0.1 C! The field of complex numbers C is the natural framework for the analysis of linear, timeinvariant filters, and the development of algorithms for circulant matrices. △ −0.2 −0.2 −0.2 −0.2 −0.3 −0.3 −0.3 −0.3 −0.4 real part imaginary part 1 2 3 4 5 6 7 8 vector component index −0.4 real part imaginary part 1 2 3 4 5 6 7 8 vector component index −0.4 real part imaginary part 1 2 3 4 5 6 7 8 vector component index −0.4 real part imaginary part 1 2 3 4 5 6 7 8 vector component index Eigenvectors of matrix C 2 ✎ notation: nth root of unity ω n := exp(−2πi/n) = cos(2π/n) − i sin(2π/n), n ∈ N vector component value 0 −0.05 −0.1 −0.15 −0.2 −0.25 −0.3 −0.35 Circulant matrix 2, eigenvector 1 vector component value 0.4 0.3 0.2 0.1 0 −0.1 −0.2 −0.3 Circulant matrix 2, eigenvector 2 vector component value 0.4 0.3 0.2 0.1 0 −0.1 −0.2 −0.3 Circulant matrix 2, eigenvector 3 vector component value 0.4 0.3 0.2 0.1 0 −0.1 −0.2 −0.3 Circulant matrix 2, eigenvector 4 Ôº º¾ satisfies ω n = ωn −1 , ωn n = 1 , ωn n/2 n−1 ∑ ωn kj = k=0 { n , if j = 0 mod n , 0 , if j ≠ 0 mod n . = −1 , ω k n = ω k+n n ∀ k ∈ Z , (7.2.1) (7.2.2) Ôº¼ º¾ −0.4 real part imaginary part 1 2 3 4 5 6 7 8 vector component index −0.4 real part imaginary part 1 2 3 4 5 6 7 8 vector component index −0.4 real part imaginary part 1 2 3 4 5 6 7 8 vector component index −0.4 real part imaginary part 1 2 3 4 5 6 7 8 vector component index
(7.2.2) is a simple consequence of the geometric sum formula n−1 ∑ q k = 1 − qn 1 − q k=0 n−1 ∑ ⇒ ωn kj = 1 − ωnj n k=0 1 − ωn j because exp(−2πij) = ω nj n = (ω n n) j = 1 for all j ∈ Z. ∀ q ∈ C \ {1} , n ∈ N . (7.2.3) = 1 − exp(−2πij) 1 − exp(−2πij/n) = 0 , Now we want to confirm the conjecture gleaned from Ex. 7.2.1 that vectors with powers of roots of unity are eigenvectors for any circulant matrix. We do this by simple and straightforward computations: Consider: C ∈ C n,n circulant matrix (→ Def. 7.1.3), c ij = u i−j , for n-periodic sequence (u k ) k∈Z , u k ∈ C v k ∈ C n with v k := (ω jk n ) n−1 j=0 ∈ Cn , k ∈ {0,...,n − 1}. Ôº½ º¾ (7.2.2) ⇒ orthogonality of basis vectors: v k := (ωn jk n−1 ) j=0 n−1 ∈ Cn : vk H v ∑ m = j=0 ωn −jk ωn jm n−1 ∑ = j=0 ω (m−k)j n (7.2.2) = 0 , if k ≠ m . (7.2.5) Matrix of change of basis trigonometrical basis → standard basis: Fourier-matrix ⎛ ωn 0 ωn 0 · · · ω 0 ⎞ n ωn 0 ωn 1 · · · ω n−1 n F n = ωn 0 ωn 2 · · · ωn 2n−2 ( = ω lj ) n−1 n ⎜ ⎝ . . . ⎟ ∈ l,j=0 Cn,n . (7.2.6) ⎠ ωn 0 ωn n−1 · · · ω n (n−1)2 ✬ Lemma 7.2.1 (Properties of Fourier matrix). ✫ The scaled Fourier-matrix 1 √ n F n is unitary (→ Def. 2.8.1): F −1 ✩ º¾ ✪ n = 1 n FH n = 1 n F n. Ôº¿ (u j−l ω lk n ) l∈Z is n-periodic! n−1 ∑ (Cv k ) j = u j−l ωn lk = l=0 n−1 ∑ = u l ω n (j−l)k l=0 j∑ l=j−n+1 = ω jk n n−1 ∑ l=0 u j−l ω lk n u l ω −lk n = λ k · ω jk n = λ k · (v k ) j . change of summation index independent of j ! (7.2.4) Proof. ( F n F H n The lemma is immediate from (7.2.5) and (7.2.2),because n−1 )l,j = ∑ k=0 ω n (l−1)k n−1 ω (j−1)k ∑ n = Remark 7.2.4 (Spectrum of Fourier matrix). k=0 ω n (l−1)k ωn −(j−1)k n−1 ∑ = k=0 1 n 2F4 n = I ⇒ σ( √ 1 n F n ) ⊂ {1, −1, i, −i} , ω k(l−j) n , 1 ≤ l, j ≤ n . v is eigenvector of C to eigenvalue λ k = ∑ n−1 l=0 u lω −lk n . because, if λ ∈ C is an eigenvalue of F n , then there is an eigenvector x ∈ C n \ {0} such that F n x = λx, see Def. 5.1.1. △ Orthogonal trigonometric basis of C n = eigenvector basis for circulant matrices ⎧⎛ ω 0 ⎞⎛ n ω 0 ⎞ ⎛ ω 0 ⎞⎛ n n ω 0 ⎞⎫ ⎪⎨ . ω 1 n ω n−2 n n ω n−1 . ω 2(n−2) n · · · n ω 2(n−1) ⎪⎬ n . ⎜ ⎟⎜ ⎟ ⎝ . ⎠⎝ . ⎠ ⎜ ⎟⎜ ⎟ ⎝ .. ⎠⎝ .. ⎠ ⎪⎩ ωn 0 ωn n−1 ω n (n−1)(n−2) ω n (n−1)2 ⎪⎭ Ôº¾ º¾ ✬ Lemma 7.2.2 (Diagonalization of circulant matrices (→ Def. 7.1.3)). For any circulant matrix C ∈ K n,n , c ij = u i−j , (u k ) k∈Z n-periodic sequence, holds true CF n = F n diag(d 1 ,...,d n ) , d = F n (u 0 ,...,u n−1 ) T . ✫ ✩ Ôº º¾ ✪
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2 Direct Methods for Linear Systems
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III Integration of Ordinary Differe
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Extra questions for course evaluati
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1.1.2 Matrices Matrices = two-dimen
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Remark 1.2.1 (Row-wise & column-wis
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1.3 Complexity/computational effort
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Syntax of BLAS calls: The functions
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4 { 5 a ssert ( this−>n==B. n &&
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34 long r t 0 ; 35 bool bStarted ;
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Obviously, left multiplication with
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❶: elimination step, ❷: backsub
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A direct way to LU-decomposition:
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Solution of LŨx = b: x ( ) 2ǫ = 1
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numerically equivalent ˆ= same res
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Code 2.4.8: Finding outeps in MATLA
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Terminology: Def. 2.5.5 introduces
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Example 2.5.5 (Instability of multi
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Note: sensitivity gauge depends on
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6 for i =1:20 7 n = 2^ i ; m = n /
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0 20 40 60 80 100 120 140 160 180 2
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Use sparse matrix format: 10 1 10 2
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Envelope-aware LU-factorization: 0
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0 20 40 60 80 100 0 20 0 20 0 20 De
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Evident: symmetry of Ã − bbT a 1
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9 ylabel ( ’ { \ b f c o n d i t
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Mapping a ∈ K n to a multiple of
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Then store G ij (a,b) as triple (i,
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Recall: e i ˆ= i-th unit vector Ch
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Computation of Choleskyfactorizatio
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1 ∃ (partial) cyclic row permutat
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Definition 3.1.3 (Local and global
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Example 3.1.6 (quadratic convergenc
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k |x (k) − π| L 1−L |x (k) −
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(x (k) ) k∈N0 Cauchy sequence ➤
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Termination criterion for contracti
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Given x (k) ∈ I, next iterate :=
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secant method ( MATLAB implementati
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Assuming p = 1: p > 1: ∥ C ∥e (
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This is a simple computation: DG(x)
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k x (k) ǫ k := ‖x ∗ − x (k)
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Code 3.4.14: Damped Newton method (
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MATLAB-CODE: Broyden method (3.4.11
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Algorithm 4.1.3 (Steepest descent).
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Example 4.1.8 (Convergence of gradi
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Page 95 and 96: Idea: Solve Ax = b approximately in
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Page 109 and 110: ✬ ✩ ✬ ✩ Lemma 5.3.4 (Ncut a
Page 111 and 112: In other words, roundoff errors may
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Page 117 and 118: ✬ Residuals r 0 ,...,r m−1 gene
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Page 123 and 124: Illustration: columns = ONB of Im(A
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Page 129 and 130: Consider the linear least squares p
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f 10 Numerical Quadrature Numerical
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f • n = 1: Trapezoidal rule • n
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For fixed local n-point quadrature
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Equidistant trapezoidal rule (order
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Heuristics: A quadrature formula ha
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20 Zeros of Legendre polynomials in
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|quadrature error| 10 0 Numerical q
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f f • line 9: estimate for global
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Model: autonomous Lotka-Volterra OD
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Example 11.1.6 (Transient circuit s
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y y 1 y(t) y 0 t t 0 t 1 Fig. 133 e
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for discrete evolution defined on I
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⇒ if y ∈ C 2 ([0, T]), then y(t
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The implementation of an s-stage ex
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Example 11.5.2 (Blow-up). ✸ toler
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However, it would be foolish not to
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0 0.2 0.4 0.6 0.8 1 1.2 1.4 1.6 1.8
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4 3 2 abstol = 0.000010, reltol = 0
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✸ ✸ Motivated by the considerat
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0 1 2 3 4 5 6 u(t),v(t) 0.01 0.008
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MATLAB-CODE : Explicit integration
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Shorthand notation for Runge-Kutta
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13 Structure Preservation 13.1 Diss
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[18] G. GOLUB AND C. VAN LOAN, Matr
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linear in Gauss-Newton method, 538
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Chebychev nodes, 678 double, 634 fo
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x∗ n y ˆ= discrete periodic conv
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Gaussian elimination with pivoting
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