Matrix functions and their approximation. Krylov subspaces

Transcription

1 [ 1 / 31 ] University of Cyprus Matrix functions and their approximation using Krylov subspaces Matrixfunktionen und ihre Approximation in Krylov-Unterräumen Stefan Güttel stefan@guettel.com Nicosia, 24th of January 2006 Matrix functions Krylov subspaces

2 [ 2 / 31 ] University of Cyprus Overview 1 Matrix functions Introduction Definitions Properties 2 Krylov subspaces Arnoldi method First error bounds Matrix functions Krylov subspaces

3 [ 3 / 31 ] University of Cyprus What is a matrix function? In general... f : D R, from a domain D C k l to some range area R C m n. f : D R m = n = 1 m = 1 or n = 1 m, n arbitrary k = l = 1 scalar function of a single variable vector function of a single variable matrix-valued f. of a single variable k = 1 or l = 1 scalar function of a vector vector field matrix function of a vector k, l arbitrary scalar function vector function matrix function Tabelle: Classification of matrix functions

4 [ 3 / 31 ] University of Cyprus What is a matrix function? In general... f : D R, from a domain D C k l to some range area R C m n. f : D R m = n = 1 m = 1 or n = 1 m, n arbitrary k = l = 1 scalar function of a single variable vector function of a single variable matrix-valued f. of a single variable k = 1 or l = 1 scalar function of a vector vector field matrix function of a vector k, l arbitrary scalar function vector function matrix function Tabelle: Classification of matrix functions

5 [ 4 / 31 ] University of Cyprus Definition 1 Polynomial matrix functions Given A C N N and p(z) of degree m with complex coefficients, i.e. p(z) = α m z m + α m 1 z m α 0. Notation: p P m (z). Since the powers I, A, A 2,... exist we may give the following Definition p(a) := α m A m + α m 1 A m α 0 I C N N. p is a polynomial matrix function. (D1) We no longer have to distinguish between P m (z) and the set of polynomials in A of degree m. We simply write P m.

6 [ 4 / 31 ] University of Cyprus Definition 1 Polynomial matrix functions Given A C N N and p(z) of degree m with complex coefficients, i.e. p(z) = α m z m + α m 1 z m α 0. Notation: p P m (z). Since the powers I, A, A 2,... exist we may give the following Definition p(a) := α m A m + α m 1 A m α 0 I C N N. p is a polynomial matrix function. (D1) We no longer have to distinguish between P m (z) and the set of polynomials in A of degree m. We simply write P m.

7 [ 4 / 31 ] University of Cyprus Definition 1 Polynomial matrix functions Given A C N N and p(z) of degree m with complex coefficients, i.e. p(z) = α m z m + α m 1 z m α 0. Notation: p P m (z). Since the powers I, A, A 2,... exist we may give the following Definition p(a) := α m A m + α m 1 A m α 0 I C N N. p is a polynomial matrix function. (D1) We no longer have to distinguish between P m (z) and the set of polynomials in A of degree m. We simply write P m.

8 [ 4 / 31 ] University of Cyprus Definition 1 Polynomial matrix functions Given A C N N and p(z) of degree m with complex coefficients, i.e. p(z) = α m z m + α m 1 z m α 0. Notation: p P m (z). Since the powers I, A, A 2,... exist we may give the following Definition p(a) := α m A m + α m 1 A m α 0 I C N N. p is a polynomial matrix function. (D1) We no longer have to distinguish between P m (z) and the set of polynomials in A of degree m. We simply write P m.

9 [ 5 / 31 ] University of Cyprus Properties of polynomials in matrices Lemma Let p P m be a polynomial, A C N N and A = TJT 1 where J = diag(j 1, J 2,..., J k ) is block-diagonal. Then 1 p(a) = Tp(J)T 1, 2 p(j) = diag (p(j 1 ), p(j 2 ),..., p(j k )), 3 If Av = λv then p(a)v = p(λ)v, 4 Given another polynomial p P m, then p(a) p(a) = p(a)p(a).

10 [ 6 / 31 ] University of Cyprus The Jordan canonical form Every square matrix A is similar to a block-diagonal Jordan matrix J = diag(j 1, J 2,..., J k ), where each Jordan block J j = J j (λ j ) C n j n j has entries λ j on the main diagonal and ones on the first upper diagonal (j = 1, 2,..., k): J j (λ j ) := toep(λ j, 1) = λ j 1 λ j λ j 1 λ j. We say J = T 1 AT is a Jordan canonical form (JCF) of A. The columns of T are the generalized eigenvectors of A.

11 [ 6 / 31 ] University of Cyprus The Jordan canonical form Every square matrix A is similar to a block-diagonal Jordan matrix J = diag(j 1, J 2,..., J k ), where each Jordan block J j = J j (λ j ) C n j n j has entries λ j on the main diagonal and ones on the first upper diagonal (j = 1, 2,..., k): J j (λ j ) := toep(λ j, 1) = λ j 1 λ j λ j 1 λ j. We say J = T 1 AT is a Jordan canonical form (JCF) of A. The columns of T are the generalized eigenvectors of A.

12 [ 7 / 31 ] University of Cyprus The Jordan canonical form Given a Jordan block J := toep(λ, 1) C n n. Let f (z) := z m be the monomial of degree m. Then ( ( ) ( ) f (J) = toep λ m, mλ m 1 m,..., λ m i m,..., )λ m min{m,n 1} i min{m, n 1} ( = toep f (λ), f (λ),..., f (i) (λ),..., f ) (min{m,n 1}) (λ) i! min{m, n 1}! ( = toep f (λ), f (λ),..., f (i) (λ),..., f ) (n 1) (λ). i! (n 1)! f (J) is already defined if f, f,..., f (n 1) exist in an open subset of C containing λ.

13 [ 7 / 31 ] University of Cyprus The Jordan canonical form Given a Jordan block J := toep(λ, 1) C n n. Let f (z) := z m be the monomial of degree m. Then ( ( ) ( ) f (J) = toep λ m, mλ m 1 m,..., λ m i m,..., )λ m min{m,n 1} i min{m, n 1} ( = toep f (λ), f (λ),..., f (i) (λ),..., f ) (min{m,n 1}) (λ) i! min{m, n 1}! ( = toep f (λ), f (λ),..., f (i) (λ),..., f ) (n 1) (λ). i! (n 1)! f (J) is already defined if f, f,..., f (n 1) exist in an open subset of C containing λ.

14 [ 7 / 31 ] University of Cyprus The Jordan canonical form Given a Jordan block J := toep(λ, 1) C n n. Let f (z) := z m be the monomial of degree m. Then ( ( ) ( ) f (J) = toep λ m, mλ m 1 m,..., λ m i m,..., )λ m min{m,n 1} i min{m, n 1} ( = toep f (λ), f (λ),..., f (i) (λ),..., f ) (min{m,n 1}) (λ) i! min{m, n 1}! ( = toep f (λ), f (λ),..., f (i) (λ),..., f ) (n 1) (λ). i! (n 1)! f (J) is already defined if f, f,..., f (n 1) exist in an open subset of C containing λ.

15 [ 7 / 31 ] University of Cyprus The Jordan canonical form Given a Jordan block J := toep(λ, 1) C n n. Let f (z) := z m be the monomial of degree m. Then ( ( ) ( ) f (J) = toep λ m, mλ m 1 m,..., λ m i m,..., )λ m min{m,n 1} i min{m, n 1} ( = toep f (λ), f (λ),..., f (i) (λ),..., f ) (min{m,n 1}) (λ) i! min{m, n 1}! ( = toep f (λ), f (λ),..., f (i) (λ),..., f ) (n 1) (λ). i! (n 1)! f (J) is already defined if f, f,..., f (n 1) exist in an open subset of C containing λ.

16 [ 7 / 31 ] University of Cyprus The Jordan canonical form Given a Jordan block J := toep(λ, 1) C n n. Let f (z) := z m be the monomial of degree m. Then ( ( ) ( ) f (J) = toep λ m, mλ m 1 m,..., λ m i m,..., )λ m min{m,n 1} i min{m, n 1} ( = toep f (λ), f (λ),..., f (i) (λ),..., f ) (min{m,n 1}) (λ) i! min{m, n 1}! ( = toep f (λ), f (λ),..., f (i) (λ),..., f ) (n 1) (λ). i! (n 1)! f (J) is already defined if f, f,..., f (n 1) exist in an open subset of C containing λ.

17 [ 8 / 31 ] University of Cyprus Definition 2 Definition Given A C N N with a Jordan canonical form J = T 1 AT, where J = diag(j 1, J 2,..., J k ) and J j = J j (λ j ) C n j n j (j = 1, 2,..., k). Let U be an open subset of C such that {λ 1, λ 2,..., λ k } U. Let f be a function f : U D C. Then f is defined on A if f (λ j ), f (λ j ),..., f (d λ 1) j (λ j ) exist, d λj := max{n i : i = 1, 2,..., k and λ i = λ j }. We set f (A) := T diag (f (J 1 ), f (J 2 ),..., f (J k )) T 1, (D2) where ( f (J j ) := toep f (λ j ), f (λ j ),..., f (i) (λ j ),..., f (n j 1) (λ j ) i! (n j 1)! ).

18 [ 9 / 31 ] University of Cyprus Remarks 1 This definition is independent of the choice of J and T. Hence, f (A) is uniquely determined. 2 d λ is the size of the largest Jordan block to eigenvalue λ. By Λ(A) := {λ 1, λ 2,..., λ k } we denote the set of the eigenvalues of A. The minimal polynomial of A is defined as ψ A (z) := (z λ) d λ. λ Λ(A) 3 For all the λ j being pairwise distinct, ψ A (z) = k (z λ j ) n j = χ A (z), j=1 where χ A (z) is the characteristic polynomial of A. Matrices with ψ A = χ A are called nonderogatory. 4 For all p P m there holds (D1) = (D2).

19 [ 10 / 31 ] University of Cyprus Polynomial interpolation Theorem 1 There holds if and only if f (A) = p(a) f (i) (λ) = p (i) (λ), λ Λ(A), i = 0, 1,..., d λ 1. (HIP) These are d := deg(ψ A ) interpolation conditions to p. 2 There exists a uniquely determined polynomial ˆp P d 1 that satisfies (HIP). ˆp is the Hermite interpolation polynomial. 3 Assumed p is another polynomial that satisfies (HIP). Then p(z) = ˆp(z) + ψ A (z)h(z) for some polynomial h(z) and ψ A the minimal polynomial of A.

20 [ 11 / 31 ] University of Cyprus Example (1) Let A = [α]. Then ψ A (z) = z α and deg(ψ A ) = 1, the most. Therefore f (A) = ˆp(A) with deg(ˆp) = 0, namely ˆp(A) = f (α)i. This is a degenerated case.

21 [ 12 / 31 ] University of Cyprus Example (2) Calculate ˆp for f (z) = exp(z) and A = J = ψ A (z) = (z 1)(z + 1) 2 z p(λ 1 ) = p(1) p(λ 2 ) = p( 1) p (λ 2 ) = p ( 1) p(λ 3 ) = p(0)! = exp(1) = e,! = exp( 1) = 1/e,! = exp( 1) = 1/e,! = exp(0) = 1. A solution is p(z) = e2 4e+5 z 4e 3 + (e 1)2 z 2e 2 + e2 +4e 7 z + 1 and there holds 4e p(a) = f (A) = exp(a). Because of deg(p) < deg(ψ A ), p = ˆp.

22 [ 12 / 31 ] University of Cyprus Example (2) Calculate ˆp for f (z) = exp(z) and A = J = ψ A (z) = (z 1)(z + 1) 2 z p(λ 1 ) = p(1) p(λ 2 ) = p( 1) p (λ 2 ) = p ( 1) p(λ 3 ) = p(0)! = exp(1) = e,! = exp( 1) = 1/e,! = exp( 1) = 1/e,! = exp(0) = 1. A solution is p(z) = e2 4e+5 z 4e 3 + (e 1)2 z 2e 2 + e2 +4e 7 z + 1 and there holds 4e p(a) = f (A) = exp(a). Because of deg(p) < deg(ψ A ), p = ˆp.

23 [ 12 / 31 ] University of Cyprus Example (2) Calculate ˆp for f (z) = exp(z) and A = J = ψ A (z) = (z 1)(z + 1) 2 z p(λ 1 ) = p(1) p(λ 2 ) = p( 1) p (λ 2 ) = p ( 1) p(λ 3 ) = p(0)! = exp(1) = e,! = exp( 1) = 1/e,! = exp( 1) = 1/e,! = exp(0) = 1. A solution is p(z) = e2 4e+5 z 4e 3 + (e 1)2 z 2e 2 + e2 +4e 7 z + 1 and there holds 4e p(a) = f (A) = exp(a). Because of deg(p) < deg(ψ A ), p = ˆp.

24 [ 12 / 31 ] University of Cyprus Example (2) Calculate ˆp for f (z) = exp(z) and A = J = ψ A (z) = (z 1)(z + 1) 2 z p(λ 1 ) = p(1) p(λ 2 ) = p( 1) p (λ 2 ) = p ( 1) p(λ 3 ) = p(0)! = exp(1) = e,! = exp( 1) = 1/e,! = exp( 1) = 1/e,! = exp(0) = 1. A solution is p(z) = e2 4e+5 z 4e 3 + (e 1)2 z 2e 2 + e2 +4e 7 z + 1 and there holds 4e p(a) = f (A) = exp(a). Because of deg(p) < deg(ψ A ), p = ˆp.

25 [ 12 / 31 ] University of Cyprus Example (2) Calculate ˆp for f (z) = exp(z) and A = J = ψ A (z) = (z 1)(z + 1) 2 z p(λ 1 ) = p(1) p(λ 2 ) = p( 1) p (λ 2 ) = p ( 1) p(λ 3 ) = p(0)! = exp(1) = e,! = exp( 1) = 1/e,! = exp( 1) = 1/e,! = exp(0) = 1. A solution is p(z) = e2 4e+5 z 4e 3 + (e 1)2 z 2e 2 + e2 +4e 7 z + 1 and there holds 4e p(a) = f (A) = exp(a). Because of deg(p) < deg(ψ A ), p = ˆp.

26 [ 13 / 31 ] University of Cyprus Remarks 1 Every function f ( ) that is defined on the spectrum of A C N N can be represented pointwise (i.e., for a concrete A) as a polynomial p(a) P d 1, d = deg(ψ A ). Or we might say, f is a field of polynomials. 2 f (A) depends only on the values of f, f,... on Λ(A). Thus f (A) and f (B) have the same polynomial representation for A and B having the same minimal polynomial (e.g. similar matrices). 3 If all Jordan blocks have size 1 1 and thus J is a diagonal matrix (e.g. for normal A) then (HIP) reduces to a Lagrange interpolation problem: f (λ) = p(λ), λ Λ(A). (LIP)

27 [ 14 / 31 ] University of Cyprus The components of A Let again ψ A (z) = λ Λ(A) (z λ)d λ denote the minimal polynomial of A, d = deg(ψ A ). Definition Define H := {ϕ λ,i (z) P d 1 : λ Λ(A), i = 0, 1,..., d λ 1} such that { ϕ (ν) 1, z = λ, i = ν; λ,i (z) = 0, otherwise, for all z Λ(A). H is the Hermite basis of P d 1 with respect to ψ A. (It has to be shown that all the ϕ λ,i are linearly independent.)

28 [ 15 / 31 ] University of Cyprus The components of A Hermite basis for example (2) 1.5 φ 1,0 φ 1,1 φ 0,0 φ 1,

29 [ 16 / 31 ] University of Cyprus Definition The components C λ,i of A are defined as C λ,i := ϕ λ,i (A). Lemma 1 {C λ,i : λ Λ(A); i = 0, 1,..., d λ 1} is a set of linearly independent matrices. 2 spectral resolution of A for f : 3 f (A) = λ Λ(A) d λ 1 i=0 f (i) (λ)c λ,i, λ Λ(A) C λ,0 = I and λ Λ(A) λc λ,0 + C λ,1 = A, 4 C λ,i C µ,j = C µ,j C λ,i. (SR)

30 [ 16 / 31 ] University of Cyprus Definition The components C λ,i of A are defined as C λ,i := ϕ λ,i (A). Lemma 1 {C λ,i : λ Λ(A); i = 0, 1,..., d λ 1} is a set of linearly independent matrices. 2 spectral resolution of A for f : 3 f (A) = λ Λ(A) d λ 1 i=0 f (i) (λ)c λ,i, λ Λ(A) C λ,0 = I and λ Λ(A) λc λ,0 + C λ,1 = A, 4 C λ,i C µ,j = C µ,j C λ,i. (SR)

31 [ 17 / 31 ] University of Cyprus Cauchy integral formula Let f (z) be analytic in a domain G and let γ be a closed path contained in G. Then the Cauchy theorem asserts f (i) (z) = i! f (ζ) dζ (CIF) 2π i (ζ z) i+1 for any z G, wind z (γ) = 1 and i = 0, 1,... γ

32 [ 18 / 31 ] University of Cyprus The resolvent of A Lemma Let A C N N and ζ Λ(A), C λ,i the components of A. There holds R ζ (A) := (ζi A) 1 = R ζ (A) is the resolvent of A to ζ. λ Λ(A) d λ 1 i=0 i! (ζ λ) i+1 C λ,i. Beweis. For ζ Λ(A), (ζi A) is invertible because N (ζi A) = {0}. The spectral resolution of A for f ζ (λ) = 1/(ζ λ), which is defined for all λ ζ, yields the desired equivalence.

33 [ 18 / 31 ] University of Cyprus The resolvent of A Lemma Let A C N N and ζ Λ(A), C λ,i the components of A. There holds R ζ (A) := (ζi A) 1 = R ζ (A) is the resolvent of A to ζ. λ Λ(A) d λ 1 i=0 i! (ζ λ) i+1 C λ,i. Beweis. For ζ Λ(A), (ζi A) is invertible because N (ζi A) = {0}. The spectral resolution of A for f ζ (λ) = 1/(ζ λ), which is defined for all λ ζ, yields the desired equivalence.

34 [ 19 / 31 ] University of Cyprus Theorem Let A C N N, γ be a closed path surrounding all λ Λ(A) once, f analytic in int(γ) and extending continuously to it, then f (A) = 1 f (ζ)(ζi A) 1 dζ = 1 f (ζ)r ζ (A)dζ. (D3) 2π i 2π i Beweis. γ By multiplying both sides of R ζ (A) by f (ζ)/(2π i) and integrating along γ we get d 1 f (ζ)(ζi A) 1 f (ζ) λ 1 i! dζ = 2π i γ γ 2π i (ζ λ) (i+1) C λ,i dζ λ Λ(A) i=0 = (CIF) = λ Λ(A) λ Λ(A) (SR) = f (A). d λ 1 i=0 d λ 1 i=0 γ ( i! 2π i γ f (i) (λ)c λ,i ) f (ζ) dζ C (ζ λ) i+1 λ,i

35 [ 19 / 31 ] University of Cyprus Theorem Let A C N N, γ be a closed path surrounding all λ Λ(A) once, f analytic in int(γ) and extending continuously to it, then f (A) = 1 f (ζ)(ζi A) 1 dζ = 1 f (ζ)r ζ (A)dζ. (D3) 2π i 2π i Beweis. γ By multiplying both sides of R ζ (A) by f (ζ)/(2π i) and integrating along γ we get d 1 f (ζ)(ζi A) 1 f (ζ) λ 1 i! dζ = 2π i γ γ 2π i (ζ λ) (i+1) C λ,i dζ λ Λ(A) i=0 = (CIF) = λ Λ(A) λ Λ(A) (SR) = f (A). d λ 1 i=0 d λ 1 i=0 γ ( i! 2π i γ f (i) (λ)c λ,i ) f (ζ) dζ C (ζ λ) i+1 λ,i

36 [ 20 / 31 ] University of Cyprus Power series Definition Let f be analytic in an open set U 0 and let f (z) = j=0 α jz j be the Taylor expansion of f in 0 with convergence radius τ (0, ]. Then f (A) is defined for every A with σ(a) < τ and there holds f (A) = j=0 α j A j = lim m m α j A j. j=0 (D4) j=0 α ja j converges ε > 0 n ε N 0 : j=n ε α j A j < ε. Assumed f has convergence radius τ (i.e., f (z) < for z < τ). Then j=n ε α j A j j=n ε α j A j, thus σ(a) A < τ is a sufficient criteria for convergence of j=0 α ja j (Taylor series converge absolutely!).

37 [ 20 / 31 ] University of Cyprus Power series Definition Let f be analytic in an open set U 0 and let f (z) = j=0 α jz j be the Taylor expansion of f in 0 with convergence radius τ (0, ]. Then f (A) is defined for every A with σ(a) < τ and there holds f (A) = j=0 α j A j = lim m m α j A j. j=0 (D4) j=0 α ja j converges ε > 0 n ε N 0 : j=n ε α j A j < ε. Assumed f has convergence radius τ (i.e., f (z) < for z < τ). Then j=n ε α j A j j=n ε α j A j, thus σ(a) A < τ is a sufficient criteria for convergence of j=0 α ja j (Taylor series converge absolutely!).

38 [ 21 / 31 ] University of Cyprus Power series Example Let f (z) = exp(z). f has convergence radius τ =. Thus f (A) is defined for every A C N N and there holds f (A) = exp(a) = j=0 A j j!.

39 [ 22 / 31 ] University of Cyprus Some facts Because of f (z) = α C f (A) = αi, f (z) = z f (A) = A, f (z) = g(z) + h(z) f (A) = g(a) + h(a) f (z) = g(z)h(z) f (A) = g(a)h(a), any rational identity in scalar functions of a complex variable will be fulfilled by the corresponding matrix function. Examples sin 2 (A) + cos 2 (A) = I, exp(ia) = cos(a) + i sin(a), (I A) 1 = I + A + A 2 + (for σ(a) < 1).

40 [ 22 / 31 ] University of Cyprus Some facts Because of f (z) = α C f (A) = αi, f (z) = z f (A) = A, f (z) = g(z) + h(z) f (A) = g(a) + h(a) f (z) = g(z)h(z) f (A) = g(a)h(a), any rational identity in scalar functions of a complex variable will be fulfilled by the corresponding matrix function. Examples sin 2 (A) + cos 2 (A) = I, exp(ia) = cos(a) + i sin(a), (I A) 1 = I + A + A 2 + (for σ(a) < 1).

41 [ 23 / 31 ] University of Cyprus Krylov subspaces Definition Problem Given A C N N, b C N, f defined on A. Calculate f (A)b! Definition The m-th Krylov (sub)space of A and b is defined by Lemma K m (A, b) = K m := span{b, Ab, A 2 b,..., A m 1 b}. There exists an index L = L(A, b) deg(ψ A ) such that K 1 (A, b) K 2 (A, b)... K L (A, b) = K L+1 (A, b) =... Moreover f (A)b K L. Matrix functions Krylov subspaces Arnoldi method First error bounds

42 [ 23 / 31 ] University of Cyprus Krylov subspaces Definition Problem Given A C N N, b C N, f defined on A. Calculate f (A)b! Definition The m-th Krylov (sub)space of A and b is defined by Lemma K m (A, b) = K m := span{b, Ab, A 2 b,..., A m 1 b}. There exists an index L = L(A, b) deg(ψ A ) such that K 1 (A, b) K 2 (A, b)... K L (A, b) = K L+1 (A, b) =... Moreover f (A)b K L. Matrix functions Krylov subspaces Arnoldi method First error bounds

43 [ 23 / 31 ] University of Cyprus Krylov subspaces Definition Problem Given A C N N, b C N, f defined on A. Calculate f (A)b! Definition The m-th Krylov (sub)space of A and b is defined by Lemma K m (A, b) = K m := span{b, Ab, A 2 b,..., A m 1 b}. There exists an index L = L(A, b) deg(ψ A ) such that K 1 (A, b) K 2 (A, b)... K L (A, b) = K L+1 (A, b) =... Moreover f (A)b K L. Matrix functions Krylov subspaces Arnoldi method First error bounds

44 [ 24 / 31 ] University of Cyprus The Arnoldi process Task: Generate an orthonormal basis of K m, m L. Algorithm v 1 := b/ b for j = 2, 3,..., m w j := Av j 1 ṽ j := w j j 1 i=1 (w j, v i )v i v j := ṽ j / ṽ j end Output: A matrix V m = [v 1, v 2,..., v m ] C N m. An unreduced upper Hessenberg matrix H m C m m. Matrix functions Krylov subspaces Arnoldi method First error bounds

45 [ 25 / 31 ] University of Cyprus Arnoldi decomposition Theorem Let m < L. There exist orthonormal vectors v 1, v 2,..., v m, v m+1 C N and an unreduced upper Hessenberg matrix H m C m m, such that AV m = V m H m + h m+1,m v m+1 e T m, where V m = [v 1, v 2,..., v m ] and h m+1,m C. For m = L there holds AV m = V m H m. N N N m m m N m H m A V m = V m + h m+1,mv m+1 Matrix functions Krylov subspaces Arnoldi method First error bounds

46 [ 26 / 31 ] University of Cyprus Arnoldi approximation Lemma Let p(z) = α m z m + + α 1 z + α 0 P m be a polynomial, 1 m < L. Then there holds p(a)b = b V m p(h m )e 1 + b α m γ m v m+1, where γ m = m j=1 h j+1,j. In particular, for p P m 1, there holds p(a)b = b V m p(h m )e 1. Definition We define the Arnoldi approximation from K m (A, b) to f (A)b as f m := b V m f (H m )e 1. Matrix functions Krylov subspaces Arnoldi method First error bounds

47 [ 26 / 31 ] University of Cyprus Arnoldi approximation Lemma Let p(z) = α m z m + + α 1 z + α 0 P m be a polynomial, 1 m < L. Then there holds p(a)b = b V m p(h m )e 1 + b α m γ m v m+1, where γ m = m j=1 h j+1,j. In particular, for p P m 1, there holds p(a)b = b V m p(h m )e 1. Definition We define the Arnoldi approximation from K m (A, b) to f (A)b as f m := b V m f (H m )e 1. Matrix functions Krylov subspaces Arnoldi method First error bounds

48 [ 27 / 31 ] University of Cyprus f (z) = exp(z), N = 500, A sparse with nz = 3106 (1.25 percent) and (0, 1)-normal-distributed entries. b full with (0, 1)-normal-distributed entries f m f(a)b m Execution speed: expm(a)*b s, f s. Matrix functions Krylov subspaces Arnoldi method First error bounds

49 [ 28 / 31 ] University of Cyprus Krylov subspace methods? Why to use them? expm(a), logm(a), funm(a,@sin), etc. operate only on full matrices, Arnoldi methods involve only matrix-vector-products Ab, speed and storage matters. Why to seek for convergence estimates? Iterative method, break condition? We only know m L, but L? In general, no residual available! Matrix functions Krylov subspaces Arnoldi method First error bounds

50 [ 28 / 31 ] University of Cyprus Krylov subspace methods? Why to use them? expm(a), logm(a), funm(a,@sin), etc. operate only on full matrices, Arnoldi methods involve only matrix-vector-products Ab, speed and storage matters. Why to seek for convergence estimates? Iterative method, break condition? We only know m L, but L? In general, no residual available! Matrix functions Krylov subspaces Arnoldi method First error bounds

51 [ 29 / 31 ] University of Cyprus How good are Krylov approximations? Remember: f m = b V m f (H m )e 1. The best approximation (for the 2 -norm) f m to f (A)b from K m (A, b) is its orthogonal projection, i.e. f m = V m V H m f (A)b = b V m [V H m f (A)V m ]e 1 which is computationally unfeasible. Another representation of f m is f m = b V m [Vm H f (V L H L VL H )V m ]e 1 = b V m [Vm H V L f (H L )VL H V m ]e 1 = b V m [I m O]f (H L )[I m O] T e 1 = b [V m O]f (H L )e 1. Furthermore f m = b V m f (V H m AV m )e 1 = b V m f ([I m O]H L )e 1. Matrix functions Krylov subspaces Arnoldi method First error bounds

52 [ 30 / 31 ] University of Cyprus How good are Krylov approximations? Lemma Let A be normal and Ω a compact set, Λ(A) Λ(H m ) Ω. Then f (A)b f m 2 2 b min max f (λ) p(λ). p P m 1 λ Ω To be continued... Matrix functions Krylov subspaces Arnoldi method First error bounds

53 [ 30 / 31 ] University of Cyprus How good are Krylov approximations? Lemma Let A be normal and Ω a compact set, Λ(A) Λ(H m ) Ω. Then f (A)b f m 2 2 b min max f (λ) p(λ). p P m 1 λ Ω To be continued... Matrix functions Krylov subspaces Arnoldi method First error bounds

54 [ 31 / 31 ] University of Cyprus Weiterführende Literatur I A. Autor. Einführung in das Präsentationswesen. Klein-Verlag, S. Jemand. On this and that. Journal of This and That, 2(1):50 100, Anhang Weiterführende Literatur