# Numerical Linear Algebra Chap. 4: Perturbation and Regularisation

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1 Numerical Linear Algebra Chap. 4: Perturbation and Regularisation Heinrich Voss Hamburg University of Technology Institute of Numerical Simulation TUHH Heinrich Voss Numerical Linear Algebra Chap. 4: Perturbation and Regularisation / 55

2 Linear systems Sensitivity of linear systems Consider the linear system of equation Ax = b (1) where A R (n,n) is a nonsingular matrix, and a perturbed system (A + A)(x + x) = b + b. (2) TUHH Heinrich Voss Numerical Linear Algebra Chap. 4: Perturbation and Regularisation / 55

3 Linear systems Sensitivity of linear systems Consider the linear system of equation Ax = b (1) where A R (n,n) is a nonsingular matrix, and a perturbed system (A + A)(x + x) = b + b. (2) Our aim is to examine how perturbations of A and of b affect the solution of the system. TUHH Heinrich Voss Numerical Linear Algebra Chap. 4: Perturbation and Regularisation / 55

4 Remarks Linear systems Small perturbations always have to be kept in mind when solving practical problems since TUHH Heinrich Voss Numerical Linear Algebra Chap. 4: Perturbation and Regularisation / 55

5 Remarks Linear systems Small perturbations always have to be kept in mind when solving practical problems since the data A and/or b may be obtained from measurements, and therefore they are erroneous TUHH Heinrich Voss Numerical Linear Algebra Chap. 4: Perturbation and Regularisation / 55

6 Remarks Linear systems Small perturbations always have to be kept in mind when solving practical problems since the data A and/or b may be obtained from measurements, and therefore they are erroneous using computers the representation of data as floating point numbers always produces errors. TUHH Heinrich Voss Numerical Linear Algebra Chap. 4: Perturbation and Regularisation / 55

7 Remarks Linear systems Small perturbations always have to be kept in mind when solving practical problems since the data A and/or b may be obtained from measurements, and therefore they are erroneous using computers the representation of data as floating point numbers always produces errors. Hence, one always has to emanate from the fact that one solves a perturbed linear system instead of the given one. However, usually the pertubations are quite small. TUHH Heinrich Voss Numerical Linear Algebra Chap. 4: Perturbation and Regularisation / 55

8 Perturbation lemma Linear systems Lemma Let B R (n,n), and assume that for some vector norm and the associate matrix norm the following inequality is satisfied B < 1. TUHH Heinrich Voss Numerical Linear Algebra Chap. 4: Perturbation and Regularisation / 55

9 Perturbation lemma Linear systems Lemma Let B R (n,n), and assume that for some vector norm and the associate matrix norm the following inequality is satisfied B < 1. Then the matrix I B is nonsingular, and it holds that (I B) B. TUHH Heinrich Voss Numerical Linear Algebra Chap. 4: Perturbation and Regularisation / 55

10 Proof Linear systems For every x R n, x 0, (I B)x x Bx x B x = (1 B ) x > 0. Therefore, the linear system (I B)x = 0 has the unique solution x = 0, and I B is nonsingular. TUHH Heinrich Voss Numerical Linear Algebra Chap. 4: Perturbation and Regularisation / 55

11 Proof Linear systems For every x R n, x 0, (I B)x x Bx x B x = (1 B ) x > 0. Therefore, the linear system (I B)x = 0 has the unique solution x = 0, and I B is nonsingular. The estimate of the norm of the inverse of I B follows from TUHH Heinrich Voss Numerical Linear Algebra Chap. 4: Perturbation and Regularisation / 55

12 Proof Linear systems For every x R n, x 0, (I B)x x Bx x B x = (1 B ) x > 0. Therefore, the linear system (I B)x = 0 has the unique solution x = 0, and I B is nonsingular. The estimate of the norm of the inverse of I B follows from 1 = (I B) 1 (I B) = (I B) 1 (I B) 1 B (I B) 1 (I B) 1 B (I B) 1 (I B) 1 B = (1 B ) (I B) 1. TUHH Heinrich Voss Numerical Linear Algebra Chap. 4: Perturbation and Regularisation / 55

13 Corollary Linear systems Let A R n be a nonsingular matrix, and A R n. Assume that A 1 A 1 for a matrix norm which is subordinate to some vector norm. TUHH Heinrich Voss Numerical Linear Algebra Chap. 4: Perturbation and Regularisation / 55

14 Corollary Linear systems Let A R n be a nonsingular matrix, and A R n. Assume that A 1 A 1 for a matrix norm which is subordinate to some vector norm. Then A + A is nonsingular, and it holds that (A + A) 1 A 1 1 A 1 A A 1 1 A 1 A TUHH Heinrich Voss Numerical Linear Algebra Chap. 4: Perturbation and Regularisation / 55

15 Proof Linear systems The existence of (A + A) 1 follows from the perturbation lemma since A < 1 A 1 1 > A 1 A A 1 A and A + A = A(I + A 1 A). TUHH Heinrich Voss Numerical Linear Algebra Chap. 4: Perturbation and Regularisation / 55

16 Proof Linear systems The existence of (A + A) 1 follows from the perturbation lemma since A < 1 A 1 1 > A 1 A A 1 A and A + A = A(I + A 1 A). (A + A) 1 = (I + A 1 A) 1 A 1 A 1 (I + A 1 A) 1 A 1 1 A 1 A A 1 1 A 1 A TUHH Heinrich Voss Numerical Linear Algebra Chap. 4: Perturbation and Regularisation / 55

17 Remark Linear systems The Corollary demonstrates that for a nonsingular matrix A the perturbed matrix A + A is also nonsingular if the perturbation A is sufficiently small. TUHH Heinrich Voss Numerical Linear Algebra Chap. 4: Perturbation and Regularisation / 55

18 Linear systems Perturbed linear system We consider the perturbed linear system (A + A)(x + x) = b + b, and we assume that the perturbation A is so small that the condition of the Corollary is satisfied. Then A + A is nonsingular. TUHH Heinrich Voss Numerical Linear Algebra Chap. 4: Perturbation and Regularisation / 55

19 Linear systems Perturbed linear system We consider the perturbed linear system (A + A)(x + x) = b + b, and we assume that the perturbation A is so small that the condition of the Corollary is satisfied. Then A + A is nonsingular. Solving for x one obtains the absolute error which is caused by the perturbations of A and b: x = (A + A) 1 ( b Ax) = (I + A 1 A) 1 A 1 ( b Ax). TUHH Heinrich Voss Numerical Linear Algebra Chap. 4: Perturbation and Regularisation / 55

20 Linear systems Perturbed linear system We consider the perturbed linear system (A + A)(x + x) = b + b, and we assume that the perturbation A is so small that the condition of the Corollary is satisfied. Then A + A is nonsingular. Solving for x one obtains the absolute error which is caused by the perturbations of A and b: x = (A + A) 1 ( b Ax) = (I + A 1 A) 1 A 1 ( b Ax). Hence, with an arbitrary vector norm and the subordinate matrix norm we obtain x (I + A 1 A) 1 A 1 ( b + A x ). TUHH Heinrich Voss Numerical Linear Algebra Chap. 4: Perturbation and Regularisation / 55

21 Linear systems Perturbed linear system ct. For b 0 and as a consequence x 0 it holds for the relative error x / x that ( ) x b x (I + A 1 A) 1 A 1 x + A. (3) TUHH Heinrich Voss Numerical Linear Algebra Chap. 4: Perturbation and Regularisation / 55

22 Linear systems Perturbed linear system ct. For b 0 and as a consequence x 0 it holds for the relative error x / x that ( ) x b x (I + A 1 A) 1 A 1 x + A. (3) and the Corollary yields x x A 1 ( 1 A 1 A b A A 1 A 1 A 1 A A A b + A ) ( A A + b ). (4) b TUHH Heinrich Voss Numerical Linear Algebra Chap. 4: Perturbation and Regularisation / 55

23 Linear systems Perturbed linear system ct. For b 0 and as a consequence x 0 it holds for the relative error x / x that ( ) x b x (I + A 1 A) 1 A 1 x + A. (3) and the Corollary yields x x A 1 ( 1 A 1 A b A A 1 A 1 A 1 A A A b + A ) ( A A + b ). (4) b Hence, for small perturbations (such that the denominator does not deviate very much from 1) the relative error b of the right hand side and the relative error A of the system matrix are amplified by the factor A 1 A. This amplification factor is called condition of the matrix A. TUHH Heinrich Voss Numerical Linear Algebra Chap. 4: Perturbation and Regularisation / 55

24 Definition Linear systems Let A C (n,n) be a nonsingular matrix, and let be a matrix norm on C (n,n) which is subordinate to some vector norm. TUHH Heinrich Voss Numerical Linear Algebra Chap. 4: Perturbation and Regularisation / 55

25 Definition Linear systems Let A C (n,n) be a nonsingular matrix, and let be a matrix norm on C (n,n) which is subordinate to some vector norm. Then κ(a) := A 1 A is called condition of the matrix A (or of the linear system of equations (1)) corresponding to the norm. TUHH Heinrich Voss Numerical Linear Algebra Chap. 4: Perturbation and Regularisation / 55

26 Definition Linear systems Let A C (n,n) be a nonsingular matrix, and let be a matrix norm on C (n,n) which is subordinate to some vector norm. Then κ(a) := A 1 A is called condition of the matrix A (or of the linear system of equations (1)) corresponding to the norm. Remark For every nonsingular matrix A and every norm it holds that κ(a) 1, because 1 = I = AA 1 A A 1 = κ(a). TUHH Heinrich Voss Numerical Linear Algebra Chap. 4: Perturbation and Regularisation / 55

27 Theorem Linear systems Let A, A R (n,n) and b, b R n, b 0, such that A is nonsingular, and assume that A 1 A < 1 for some matrix norm which is subordinate to some vector norm. TUHH Heinrich Voss Numerical Linear Algebra Chap. 4: Perturbation and Regularisation / 55

28 Theorem Linear systems Let A, A R (n,n) and b, b R n, b 0, such that A is nonsingular, and assume that A 1 A < 1 for some matrix norm which is subordinate to some vector norm. Let x and x + x be the solution of the linear system (1) and the perturbed system (2), respectively, and the following estimation of the relative error holds x x κ(a) ( A 1 κ(a) A A A where κ(a) := A A 1 denotes the condition of A. + b ). b TUHH Heinrich Voss Numerical Linear Algebra Chap. 4: Perturbation and Regularisation / 55

29 Remark Linear systems Assume that the length of the mantissa (i.e. the number of leading digits in floating point representation) of our computer is l. Then that the relative input data error of A and b is 5 10 l. TUHH Heinrich Voss Numerical Linear Algebra Chap. 4: Perturbation and Regularisation / 55

30 Remark Linear systems Assume that the length of the mantissa (i.e. the number of leading digits in floating point representation) of our computer is l. Then that the relative input data error of A and b is 5 10 l. If κ(a) = 10 γ, then (not considering the round of errors which occur in the numerical method for solving the linear system) we have to expect a relative error of approximately 5 10 γ l for a numerical solution the linear system Ax = b. TUHH Heinrich Voss Numerical Linear Algebra Chap. 4: Perturbation and Regularisation / 55

31 Remark Linear systems Assume that the length of the mantissa (i.e. the number of leading digits in floating point representation) of our computer is l. Then that the relative input data error of A and b is 5 10 l. If κ(a) = 10 γ, then (not considering the round of errors which occur in the numerical method for solving the linear system) we have to expect a relative error of approximately 5 10 γ l for a numerical solution the linear system Ax = b. Roughly speaking solving a linear system numerically we are loosing γ digits Stellen if the order of magnitude of the condition of the system matrix A is 10 γ. This loss of accuracy has nothing to do with the algorithm of choice. It is problem immanent. TUHH Heinrich Voss Numerical Linear Algebra Chap. 4: Perturbation and Regularisation / 55

32 Example Linear systems Consider the linear system of equations ( ) 1 1 x = which obviously has the solution x = (1, 1) T. ( ) 2, TUHH Heinrich Voss Numerical Linear Algebra Chap. 4: Perturbation and Regularisation / 55

33 Example Linear systems Consider the linear system of equations ( ) 1 1 x = which obviously has the solution x = (1, 1) T. ( ) 2, For x + x := (5, 3.002) T it holds that ( ) A(x + x) = =: b + b TUHH Heinrich Voss Numerical Linear Algebra Chap. 4: Perturbation and Regularisation / 55

34 Example Linear systems Consider the linear system of equations ( ) 1 1 x = which obviously has the solution x = (1, 1) T. ( ) 2, For x + x := (5, 3.002) T it holds that ( ) A(x + x) = =: b + b Hence, b b = and x x = 4.002, TUHH Heinrich Voss Numerical Linear Algebra Chap. 4: Perturbation and Regularisation / 55

35 Example ct. Linear systems and it follows for the condition κ (A) = TUHH Heinrich Voss Numerical Linear Algebra Chap. 4: Perturbation and Regularisation / 55

36 Example ct. Linear systems and it follows for the condition κ (A) = Indeed and therefore A 1 = ( 999 ) κ (A) = TUHH Heinrich Voss Numerical Linear Algebra Chap. 4: Perturbation and Regularisation / 55

37 Example ct. Linear systems and it follows for the condition κ (A) = Indeed and therefore A 1 = ( 999 ) κ (A) = This example demonstrates that the estimation of the relative error of the solution of a perturbed system is sharp. TUHH Heinrich Voss Numerical Linear Algebra Chap. 4: Perturbation and Regularisation / 55

38 Linear systems Geometric condition The following Theorem contains a geometric characterization of the condition number. It says that the relative distance of a nonsingular matrix to the closest singular matrix with respect to Euclidean norm is the reciprokal of the condition number. TUHH Heinrich Voss Numerical Linear Algebra Chap. 4: Perturbation and Regularisation / 55

39 Linear systems Geometric condition The following Theorem contains a geometric characterization of the condition number. It says that the relative distance of a nonsingular matrix to the closest singular matrix with respect to Euclidean norm is the reciprokal of the condition number. Theorem Let A R (n,n) be nonsingular. Then it holds that { A 2 min A 2 } : A + A singular} = 1 κ 2 (A). TUHH Heinrich Voss Numerical Linear Algebra Chap. 4: Perturbation and Regularisation / 55

40 Proof Linear systems It suffices to prove that min { A 2 : A + A singular} = 1/ A 1 2. TUHH Heinrich Voss Numerical Linear Algebra Chap. 4: Perturbation and Regularisation / 55

41 Proof Linear systems It suffices to prove that min { A 2 : A + A singular} = 1/ A 1 2. That the minimum is at least 1/ A 1 2 follows from the perturbation lemma: for A 2 < 1/ A 1 2 it holds that 1 > A 2 A 1 2 A 1 A 2. TUHH Heinrich Voss Numerical Linear Algebra Chap. 4: Perturbation and Regularisation / 55

42 Proof Linear systems It suffices to prove that min { A 2 : A + A singular} = 1/ A 1 2. That the minimum is at least 1/ A 1 2 follows from the perturbation lemma: for A 2 < 1/ A 1 2 it holds that 1 > A 2 A 1 2 A 1 A 2. Hence, and is invertible. I + A 1 A = A 1 (A + A), A + A TUHH Heinrich Voss Numerical Linear Algebra Chap. 4: Perturbation and Regularisation / 55

43 Proof ct. Linear systems We now construct a matrix A, such that A + Ais singular and A 2 = 1/ A 1 2 which demonstrates that the minimum is greater or equal to 1/ A 1 2. TUHH Heinrich Voss Numerical Linear Algebra Chap. 4: Perturbation and Regularisation / 55

44 Proof ct. Linear systems We now construct a matrix A, such that A + Ais singular and A 2 = 1/ A 1 2 which demonstrates that the minimum is greater or equal to 1/ A 1 2. From A 1 2 = max x 0 A 1 x 2 x 2 it follows that there exists x satisfying x 2 = 1 and A 1 2 = A 1 x 2 > 0. TUHH Heinrich Voss Numerical Linear Algebra Chap. 4: Perturbation and Regularisation / 55

45 Proof ct. Linear systems We now construct a matrix A, such that A + Ais singular and A 2 = 1/ A 1 2 which demonstrates that the minimum is greater or equal to 1/ A 1 2. From A 1 2 = max x 0 A 1 x 2 x 2 it follows that there exists x satisfying x 2 = 1 and A 1 2 = A 1 x 2 > 0. With this x we define y := A 1 x A 1 = A 1 x x 2 A 1 and A := xy T 2 A 1. 2 TUHH Heinrich Voss Numerical Linear Algebra Chap. 4: Perturbation and Regularisation / 55

46 Proof ct. Linear systems Then it holds that y 2 = 1 and A 2 = max z 0 xy T z 2 A 1 = max 2 z 2 z 0 where the maximum is attained for z = y, e.g. y T z z 2 x 2 A 1 2 = 1 A 1 2, TUHH Heinrich Voss Numerical Linear Algebra Chap. 4: Perturbation and Regularisation / 55

47 Proof ct. Linear systems Then it holds that y 2 = 1 and A 2 = max z 0 xy T z 2 A 1 = max 2 z 2 z 0 where the maximum is attained for z = y, e.g. y T z z 2 x 2 A 1 2 = 1 A 1 2, From (A + A)y = Ay xy T y A 1 2 = we obtain the singularity of A + A. x A 1 2 x A 1 2 = 0 TUHH Heinrich Voss Numerical Linear Algebra Chap. 4: Perturbation and Regularisation / 55

48 Theorem Least squares problems Let A = UΣV H be the singular value decomposition of A R m n where σ 1 σ 2 σ r > σ r+1 = = σ min(m,n) = 0. Then it holds that TUHH Heinrich Voss Numerical Linear Algebra Chap. 4: Perturbation and Regularisation / 55

49 Theorem Least squares problems Let A = UΣV H be the singular value decomposition of A R m n where σ 1 σ 2 σ r > σ r+1 = = σ min(m,n) = 0. Then it holds that (i) rank(a) = r, TUHH Heinrich Voss Numerical Linear Algebra Chap. 4: Perturbation and Regularisation / 55

50 Theorem Least squares problems Let A = UΣV H be the singular value decomposition of A R m n where σ 1 σ 2 σ r > σ r+1 = = σ min(m,n) = 0. Then it holds that (i) rank(a) = r, (ii) null(a) := {x R n : Ax = 0} = span{v r+1,..., v n }, TUHH Heinrich Voss Numerical Linear Algebra Chap. 4: Perturbation and Regularisation / 55

51 Theorem Least squares problems Let A = UΣV H be the singular value decomposition of A R m n where σ 1 σ 2 σ r > σ r+1 = = σ min(m,n) = 0. Then it holds that (i) rank(a) = r, (ii) null(a) := {x R n : Ax = 0} = span{v r+1,..., v n }, (iii) range(a) := {Ax : x R n } = span{u 1,..., u r }, TUHH Heinrich Voss Numerical Linear Algebra Chap. 4: Perturbation and Regularisation / 55

52 Theorem Least squares problems Let A = UΣV H be the singular value decomposition of A R m n where σ 1 σ 2 σ r > σ r+1 = = σ min(m,n) = 0. Then it holds that (i) rank(a) = r, (ii) null(a) := {x R n : Ax = 0} = span{v r+1,..., v n }, (iii) range(a) := {Ax : x R n } = span{u 1,..., u r }, (iv) A = r i=1 σ i u i (v i ) T = U r Σ r V T r with U r = (u 1,..., u r ), V r = (v 1,..., v r ), Σ r = diag(σ 1,..., σ r ), TUHH Heinrich Voss Numerical Linear Algebra Chap. 4: Perturbation and Regularisation / 55

53 Theorem Least squares problems Let A = UΣV H be the singular value decomposition of A R m n where σ 1 σ 2 σ r > σ r+1 = = σ min(m,n) = 0. Then it holds that (i) rank(a) = r, (ii) null(a) := {x R n : Ax = 0} = span{v r+1,..., v n }, (iii) range(a) := {Ax : x R n } = span{u 1,..., u r }, (iv) A = r i=1 (v) A 2 S := m σ i u i (v i ) T = U r Σ r V T r with U r = (u 1,..., u r ), V r = (v 1,..., v r ), Σ r = diag(σ 1,..., σ r ), i=1 j=1 n aij 2 = r σi 2, i=1 TUHH Heinrich Voss Numerical Linear Algebra Chap. 4: Perturbation and Regularisation / 55

54 Theorem Least squares problems Let A = UΣV H be the singular value decomposition of A R m n where σ 1 σ 2 σ r > σ r+1 = = σ min(m,n) = 0. Then it holds that (i) rank(a) = r, (ii) null(a) := {x R n : Ax = 0} = span{v r+1,..., v n }, (iii) range(a) := {Ax : x R n } = span{u 1,..., u r }, (iv) A = r i=1 (v) A 2 S := m (vi) A 2 := σ 1. σ i u i (v i ) T = U r Σ r V T r with U r = (u 1,..., u r ), V r = (v 1,..., v r ), Σ r = diag(σ 1,..., σ r ), i=1 j=1 n aij 2 = r σi 2, i=1 TUHH Heinrich Voss Numerical Linear Algebra Chap. 4: Perturbation and Regularisation / 55

55 Proof Least squares problems (i): Multiplication by nonsingular matrices U T und V does not change the rank of A. Therefore, rank(a) = rank(σ) = r. TUHH Heinrich Voss Numerical Linear Algebra Chap. 4: Perturbation and Regularisation / 55

56 Proof Least squares problems (i): Multiplication by nonsingular matrices U T und V does not change the rank of A. Therefore, rank(a) = rank(σ) = r. (ii): From V T v i = e i it follows that Av i = UΣV T v i = UΣe i = 0 for i = r + 1,..., n. Hence, v r+1,..., v n null(a). TUHH Heinrich Voss Numerical Linear Algebra Chap. 4: Perturbation and Regularisation / 55

57 Proof Least squares problems (i): Multiplication by nonsingular matrices U T und V does not change the rank of A. Therefore, rank(a) = rank(σ) = r. (ii): From V T v i = e i it follows that Av i = UΣV T v i = UΣe i = 0 for i = r + 1,..., n. Hence, v r+1,..., v n null(a). dim null(a) = n r implies that these vectors form a basis of null(a). TUHH Heinrich Voss Numerical Linear Algebra Chap. 4: Perturbation and Regularisation / 55

58 Proof ct. Least squares problems (iii): From A = UΣV T we obtain Range(A) = U Range(Σ) = U span(e 1,..., e r ) = span(u 1,..., u r ). TUHH Heinrich Voss Numerical Linear Algebra Chap. 4: Perturbation and Regularisation / 55

59 Proof ct. Least squares problems (iii): From A = UΣV T we obtain Range(A) = U Range(Σ) = U span(e 1,..., e r ) = span(u 1,..., u r ). (iv): Blockmatrix multiplication yields A = UΣV T = ( (v 1 ) T u 1... u m) Σ. = (v n ) T r σ i u i (v i ) T. i=1 TUHH Heinrich Voss Numerical Linear Algebra Chap. 4: Perturbation and Regularisation / 55

60 Proof ct. Least squares problems (v): Let A = (a 1,..., a n ). TUHH Heinrich Voss Numerical Linear Algebra Chap. 4: Perturbation and Regularisation / 55

61 Proof ct. Least squares problems (v): Let A = (a 1,..., a n ). Multiplication by the orthogonal matrix U T does not change the Euclidean length of a vector. Jence, A 2 S = n a i 2 2 = i=1 n U T a i 2 2 = U T A 2 S. i=1 TUHH Heinrich Voss Numerical Linear Algebra Chap. 4: Perturbation and Regularisation / 55

62 Proof ct. Least squares problems (v): Let A = (a 1,..., a n ). Multiplication by the orthogonal matrix U T does not change the Euclidean length of a vector. Jence, A 2 S = n a i 2 2 = i=1 n U T a i 2 2 = U T A 2 S. i=1 Similarly, multiplying the rows of U T A by the orthogonal matrix V from the right does not change their length, from which we get A 2 S = UT ΣV 2 S = Σ 2 S = r σi 2. i=1 TUHH Heinrich Voss Numerical Linear Algebra Chap. 4: Perturbation and Regularisation / 55

63 Proof ct. Least squares problems (v): Let A = (a 1,..., a n ). Multiplication by the orthogonal matrix U T does not change the Euclidean length of a vector. Jence, A 2 S = n a i 2 2 = i=1 n U T a i 2 2 = U T A 2 S. i=1 Similarly, multiplying the rows of U T A by the orthogonal matrix V from the right does not change their length, from which we get A 2 S = UT ΣV 2 S = Σ 2 S = r σi 2. i=1 (vi): A 2 is a singular value of A, i.e. A 2 σ 1 (cf. proof of the existence theorem of the SVD). Thus A 2 = max{ Ax 2 : x 2 = 1} Av 1 2 = σ 1. TUHH Heinrich Voss Numerical Linear Algebra Chap. 4: Perturbation and Regularisation / 55

64 Least squares problems Condition of a matrix Let A = UΣV T be the SVD of a nonsingular matrix A. Then A 1 = V Σ 1 U T is the SVD of A 1, from which we get A 2 = σ 1 and A 1 2 = 1 σ n. TUHH Heinrich Voss Numerical Linear Algebra Chap. 4: Perturbation and Regularisation / 55

65 Least squares problems Condition of a matrix Let A = UΣV T be the SVD of a nonsingular matrix A. Then A 1 = V Σ 1 U T is the SVD of A 1, from which we get A 2 = σ 1 and A 1 2 = 1 σ n. Hence, the condition of A with respect to the Euclidean norm is κ 2 (A) := σ 1 σ n. TUHH Heinrich Voss Numerical Linear Algebra Chap. 4: Perturbation and Regularisation / 55

66 Remark Least squares problems Let A R (n,n) have eigenvalues µ 1,..., µ n. Then it follows from Ax i = µ i x i µ i 2 = (Ax i ) H Ax i (x i ) H x i = (x i ) H A T Ax i (x i ) H x i. TUHH Heinrich Voss Numerical Linear Algebra Chap. 4: Perturbation and Regularisation / 55

67 Remark Least squares problems Let A R (n,n) have eigenvalues µ 1,..., µ n. Then it follows from Ax i = µ i x i µ i 2 = (Ax i ) H Ax i (x i ) H x i = (x i ) H A T Ax i (x i ) H x i. Rayleigh s principle yields λ min x H A T Ax x H x λ max für alle x C n, x 0, where λ min and λ max is the minimal und maximal eigenvalue of A T A, respectively. Hence, σ 1 µ i σ n for every i. TUHH Heinrich Voss Numerical Linear Algebra Chap. 4: Perturbation and Regularisation / 55

68 Remark Least squares problems Let A R (n,n) have eigenvalues µ 1,..., µ n. Then it follows from Ax i = µ i x i µ i 2 = (Ax i ) H Ax i (x i ) H x i = (x i ) H A T Ax i (x i ) H x i. Rayleigh s principle yields λ min x H A T Ax x H x λ max für alle x C n, x 0, where λ min and λ max is the minimal und maximal eigenvalue of A T A, respectively. Hence, σ 1 µ i σ n for every i. For symmetric A it folds that σ 1 = µ 1 and σ r = µ r. For non symmetric matrices this is in general not the case. TUHH Heinrich Voss Numerical Linear Algebra Chap. 4: Perturbation and Regularisation / 55

69 Least squares problems Numerical computation The singular values of A are the square roots of the eigenvalues of A T A. Hence, in principle the SVD of A can be determined with any eigensolver. TUHH Heinrich Voss Numerical Linear Algebra Chap. 4: Perturbation and Regularisation / 55

70 Least squares problems Numerical computation The singular values of A are the square roots of the eigenvalues of A T A. Hence, in principle the SVD of A can be determined with any eigensolver. To this end one has to evaluate A T A and AA T which is costly and which deteriorates the condition number considerably. TUHH Heinrich Voss Numerical Linear Algebra Chap. 4: Perturbation and Regularisation / 55

71 Least squares problems Numerical computation The singular values of A are the square roots of the eigenvalues of A T A. Hence, in principle the SVD of A can be determined with any eigensolver. To this end one has to evaluate A T A and AA T which is costly and which deteriorates the condition number considerably. Actually, one uses an algorithm of Golub and Reinsch (1971), which takes advantage of the QR algorithm for computing the eigenvalues of A T A, but which avoids the explicit computation of A T A and AA T. TUHH Heinrich Voss Numerical Linear Algebra Chap. 4: Perturbation and Regularisation / 55

72 Least squares problems Data compression The singular value decomposition can be used for data compression. This is based upon the following theorem: TUHH Heinrich Voss Numerical Linear Algebra Chap. 4: Perturbation and Regularisation / 55

73 Least squares problems Data compression The singular value decomposition can be used for data compression. This is based upon the following theorem: Theorem Let A = UΣV T be the singular value decomposition of A R m n, and let U = (u 1,..., u m ) and V = (v 1,..., v n ). TUHH Heinrich Voss Numerical Linear Algebra Chap. 4: Perturbation and Regularisation / 55

74 Least squares problems Data compression The singular value decomposition can be used for data compression. This is based upon the following theorem: Theorem Let A = UΣV T be the singular value decomposition of A R m n, and let U = (u 1,..., u m ) and V = (v 1,..., v n ). Then for k < n A k := k σ j u j (v j ) T j=1 is the best approximation of A with rank(a k ) = k with respect to the spectral norm, and it holds that A A k 2 = σ k+1. TUHH Heinrich Voss Numerical Linear Algebra Chap. 4: Perturbation and Regularisation / 55

75 Proof Least squares problems It holds that A A k 2 = n j=k+1 σ j u j (v j ) T 2 = Udiag{0,..., 0, σ k+1,..., σ n }V T 2 = σ k+1, and it remains to show, that there does not exist a matrix of rank k, the distance to A of which is less than σ k+1. TUHH Heinrich Voss Numerical Linear Algebra Chap. 4: Perturbation and Regularisation / 55

76 Proof Least squares problems It holds that A A k 2 = n j=k+1 σ j u j (v j ) T 2 = Udiag{0,..., 0, σ k+1,..., σ n }V T 2 = σ k+1, and it remains to show, that there does not exist a matrix of rank k, the distance to A of which is less than σ k+1. Let B be any matrix with rank(b) = k. Then the dimension of the null space of B is n k. The dimension of span{v 1,..., v k+1 } is k + 1, and therefore the intersection of these two spaces contains a nontrivial vector w with w 2 = 1. TUHH Heinrich Voss Numerical Linear Algebra Chap. 4: Perturbation and Regularisation / 55

77 Proof Least squares problems It holds that A A k 2 = n j=k+1 σ j u j (v j ) T 2 = Udiag{0,..., 0, σ k+1,..., σ n }V T 2 = σ k+1, and it remains to show, that there does not exist a matrix of rank k, the distance to A of which is less than σ k+1. Let B be any matrix with rank(b) = k. Then the dimension of the null space of B is n k. The dimension of span{v 1,..., v k+1 } is k + 1, and therefore the intersection of these two spaces contains a nontrivial vector w with w 2 = 1. Hence, A B 2 2 (A B)w 2 2 = Aw 2 2 = UΣV T w 2 2 = Σ(V T w) 2 2 σk+1 V 2 T w 2 2 = σk+1. 2 TUHH Heinrich Voss Numerical Linear Algebra Chap. 4: Perturbation and Regularisation / 55

78 Least squares problems Data compression Let A R (m,n) be a matrix the elements a ij of which are color values of pixels of a picture. TUHH Heinrich Voss Numerical Linear Algebra Chap. 4: Perturbation and Regularisation / 55

79 Least squares problems Data compression Let A R (m,n) be a matrix the elements a ij of which are color values of pixels of a picture. If A = UΣV T is the singular value decomposition of A, then A k = k σ j u j (v j ) T, k = 1,..., min(n, m) j=1 is an approximation to A. The storage of A k requires only k (n + m + 1)/(n m) memory cells whereas A requires mn. TUHH Heinrich Voss Numerical Linear Algebra Chap. 4: Perturbation and Regularisation / 55

80 Least squares problems Data compression Let A R (m,n) be a matrix the elements a ij of which are color values of pixels of a picture. If A = UΣV T is the singular value decomposition of A, then A k = k σ j u j (v j ) T, k = 1,..., min(n, m) j=1 is an approximation to A. The storage of A k requires only k (n + m + 1)/(n m) memory cells whereas A requires mn. Notice that that using the SVD in this manner is a very simple way of data compression. There are algorithm in image processing which are much less costly. TUHH Heinrich Voss Numerical Linear Algebra Chap. 4: Perturbation and Regularisation / 55

81 Example Least squares problems Original Figure: Original TUHH Heinrich Voss Numerical Linear Algebra Chap. 4: Perturbation and Regularisation / 55

82 Example ct. Least squares problems k=5; 2.6% Figure: Compression: k = 5 TUHH Heinrich Voss Numerical Linear Algebra Chap. 4: Perturbation and Regularisation / 55

83 Example ct. Least squares problems k=10; 5.3% Figure: Compression: k = 10 TUHH Heinrich Voss Numerical Linear Algebra Chap. 4: Perturbation and Regularisation / 55

84 Example ct. Least squares problems k=20; 10.5% Figure: Compression: k = 20 TUHH Heinrich Voss Numerical Linear Algebra Chap. 4: Perturbation and Regularisation / 55

85 Pseudoinverse Least squares problems Consider the linear least squares problem Let A R (m,n) and b R m with m n. Find x R n such that Ax b 2 = min! (1) We examine this problem taking advantage of the singular value decomposition. TUHH Heinrich Voss Numerical Linear Algebra Chap. 4: Perturbation and Regularisation / 55

86 Pseudoinverse Least squares problems Consider the linear least squares problem Let A R (m,n) and b R m with m n. Find x R n such that Ax b 2 = min! (1) We examine this problem taking advantage of the singular value decomposition. In the following we denote with σ 1 σ 2 σ r > 0 = σ r+1 = = σ n = 0 the singular values of A. A = UΣV T is the singular value decomposition of A, and u j and v k are the left and right singular vectors, respectively, i.e. the columns of U and V. TUHH Heinrich Voss Numerical Linear Algebra Chap. 4: Perturbation and Regularisation / 55

87 Least squares problems Pseudoinverse ct. Theorem Let c := U T b R m. The set of solutions of the linear least squares problem is L = x + null(a), (2) where x is the following particular solution (1): x := r i=1 c i σ i v i. (3) TUHH Heinrich Voss Numerical Linear Algebra Chap. 4: Perturbation and Regularisation / 55

88 Pseudoinverse Least squares problems Multiplying a vector by an orthogonal matrix does not change its length. Hence, with z := V T x it holds that Ax b 2 2 = U T (Ax b) 2 2 = ΣV T x U T b 2 2 = Σz c 2 2 = (σ 1 z 1 c 1,..., σ r z r c r, c r+1,..., c m ) T 2 2. TUHH Heinrich Voss Numerical Linear Algebra Chap. 4: Perturbation and Regularisation / 55

89 Pseudoinverse Least squares problems Multiplying a vector by an orthogonal matrix does not change its length. Hence, with z := V T x it holds that Ax b 2 2 = U T (Ax b) 2 2 = ΣV T x U T b 2 2 = Σz c 2 2 = (σ 1 z 1 c 1,..., σ r z r c r, c r+1,..., c m ) T 2 2. Therefore, the solution of problem (1) reads: z i := c i σ i, i = 1,..., r, und z i R beliebig für i = r + 1,..., n, i.e. x = r i=1 c i σ i v i + n i=r+1 z i v i, z i R, i = r + 1,..., n. (4) TUHH Heinrich Voss Numerical Linear Algebra Chap. 4: Perturbation and Regularisation / 55

90 Pseudoinverse Least squares problems Multiplying a vector by an orthogonal matrix does not change its length. Hence, with z := V T x it holds that Ax b 2 2 = U T (Ax b) 2 2 = ΣV T x U T b 2 2 = Σz c 2 2 = (σ 1 z 1 c 1,..., σ r z r c r, c r+1,..., c m ) T 2 2. Therefore, the solution of problem (1) reads: z i := c i σ i, i = 1,..., r, und z i R beliebig für i = r + 1,..., n, i.e. x = r i=1 c i σ i v i + n i=r+1 z i v i, z i R, i = r + 1,..., n. (4) Since the tailing n r columns of V span the null space of A, the set L of solutions of problem (1) has the form (2), (3). TUHH Heinrich Voss Numerical Linear Algebra Chap. 4: Perturbation and Regularisation / 55

91 Least squares problems Pseudonormal solution This theorem demonstrates again that the linear least squares problem (1) has a unique solution if and only if r = rank(a) = n. We enforce the uniqueness also in the case r < n requiring additionally that the Euclidean norm of the solution is minimal. TUHH Heinrich Voss Numerical Linear Algebra Chap. 4: Perturbation and Regularisation / 55

92 Least squares problems Pseudonormal solution This theorem demonstrates again that the linear least squares problem (1) has a unique solution if and only if r = rank(a) = n. We enforce the uniqueness also in the case r < n requiring additionally that the Euclidean norm of the solution is minimal. Definition Let L be the solution set of the linear least squares problem (1). x L is called pseudonormal solution of (1), if x 2 x 2 for every x L. TUHH Heinrich Voss Numerical Linear Algebra Chap. 4: Perturbation and Regularisation / 55

93 Least squares problems Pseudonormal solution ct. The representation (4) of the general solution of (1) yields that x in (3) is the pseudonormal solution of (1): n n x + z i v i 2 2 = x z i 2 v i 2 2 x 2 2. i=r+1 i=r+1 TUHH Heinrich Voss Numerical Linear Algebra Chap. 4: Perturbation and Regularisation / 55

94 Least squares problems Pseudonormal solution ct. The representation (4) of the general solution of (1) yields that x in (3) is the pseudonormal solution of (1): n n x + z i v i 2 2 = x z i 2 v i 2 2 x 2 2. i=r+1 i=r+1 The pseudonormal solution is unique, and x obviously is the only solution of (1) with x null(a) L. Hence, we obtained TUHH Heinrich Voss Numerical Linear Algebra Chap. 4: Perturbation and Regularisation / 55

95 Least squares problems Pseudonormal solution ct. The representation (4) of the general solution of (1) yields that x in (3) is the pseudonormal solution of (1): n n x + z i v i 2 2 = x z i 2 v i 2 2 x 2 2. i=r+1 i=r+1 The pseudonormal solution is unique, and x obviously is the only solution of (1) with x null(a) L. Hence, we obtained Satz There exists a unique pseudonormal solution x of problem (1) which is characterized by x null(a) L. TUHH Heinrich Voss Numerical Linear Algebra Chap. 4: Perturbation and Regularisation / 55

96 Pseudoinverse Least squares problems For every A R (m,n) R m b x R n : A x b 2 Ax b 2 x R n, x 2 minimal defines a mapping which obviously is linear (cf. the representation of x in (3)). Therefore, it can be representated as a matrix A R (n,m). TUHH Heinrich Voss Numerical Linear Algebra Chap. 4: Perturbation and Regularisation / 55

97 Pseudoinverse Least squares problems For every A R (m,n) R m b x R n : A x b 2 Ax b 2 x R n, x 2 minimal defines a mapping which obviously is linear (cf. the representation of x in (3)). Therefore, it can be representated as a matrix A R (n,m). Definition For A R (m,n) the matrix A R (n,m), such that x := A b for every b R m is the pseudonormal solution of the linear least squares problem (1) is called pseudo inverse ( or Moore-Penrose inverse) of A. TUHH Heinrich Voss Numerical Linear Algebra Chap. 4: Perturbation and Regularisation / 55

98 Pseudo inverse Least squares problems If rank(a) = n and m n, then the least squares problem (1) is uniquely solvable, and it follows from the normal equations that the solution is x = (A T A) 1 A T b. Hence, in this case A = (A T A) 1 A T. TUHH Heinrich Voss Numerical Linear Algebra Chap. 4: Perturbation and Regularisation / 55

99 Pseudo inverse Least squares problems If rank(a) = n and m n, then the least squares problem (1) is uniquely solvable, and it follows from the normal equations that the solution is x = (A T A) 1 A T b. Hence, in this case A = (A T A) 1 A T. If n = m and A is nonsingular, then it holds that A = A 1. Hence, the pseudo inverse is the usual inverse, if this one exists, and the pseudo inverse is consistent extension of the inverse. TUHH Heinrich Voss Numerical Linear Algebra Chap. 4: Perturbation and Regularisation / 55

100 Least squares problems Pseudo inverse ct. Theorem let A R (m,n) and A = UΣV T, its singular value decomposition Σ = (σ i δ ij ) i,j TUHH Heinrich Voss Numerical Linear Algebra Chap. 4: Perturbation and Regularisation / 55

101 Least squares problems Pseudo inverse ct. Theorem let A R (m,n) and A = UΣV T, its singular value decomposition Σ = (σ i δ ij ) i,j Then it holds that (i) Σ = (τ i δ ij ) j,i, τ i = { σ 1 i, falls σ i 0 0, falls σ i = 0, TUHH Heinrich Voss Numerical Linear Algebra Chap. 4: Perturbation and Regularisation / 55

102 Least squares problems Pseudo inverse ct. Theorem let A R (m,n) and A = UΣV T, its singular value decomposition Σ = (σ i δ ij ) i,j Then it holds that (i) Σ = (τ i δ ij ) j,i, τ i = (ii) A = V Σ U T. { σ 1 i, falls σ i 0 0, falls σ i = 0, TUHH Heinrich Voss Numerical Linear Algebra Chap. 4: Perturbation and Regularisation / 55

103 Least squares problems Pseudo inverse ct. Remark The explicit representation of the pseudo inverse is needed only for theoretical considerations and is never computed explicitly (similarly to the inverse of a nonsingular matrix). TUHH Heinrich Voss Numerical Linear Algebra Chap. 4: Perturbation and Regularisation / 55

104 Least squares problems Pseudo inverse ct. Remark The explicit representation of the pseudo inverse is needed only for theoretical considerations and is never computed explicitly (similarly to the inverse of a nonsingular matrix). Corollary For every matrix A R (m,n) it holds THAT A = A and (A ) T = (A T ). TUHH Heinrich Voss Numerical Linear Algebra Chap. 4: Perturbation and Regularisation / 55

105 Least squares problems Pseudo inverse ct. Remark The explicit representation of the pseudo inverse is needed only for theoretical considerations and is never computed explicitly (similarly to the inverse of a nonsingular matrix). Corollary For every matrix A R (m,n) it holds THAT A = A and (A ) T = (A T ). A has the well known properties of the inverse A 1 of a nonsingular matrix A with the only exception that in general (AB) B A. TUHH Heinrich Voss Numerical Linear Algebra Chap. 4: Perturbation and Regularisation / 55

106 Example Least squares problems Let A = ( ) 1 1 = I 0 0 ( ) ( ) 1 1 2, 1 1 TUHH Heinrich Voss Numerical Linear Algebra Chap. 4: Perturbation and Regularisation / 55

107 Example Least squares problems Let A = ( ) 1 1 = I 0 0 ( ) ( ) 1 1 2, 1 1 Its pseudo inverse is A = 1 ( ) ( ) 1 1 1/ 2 0 I = ( ) TUHH Heinrich Voss Numerical Linear Algebra Chap. 4: Perturbation and Regularisation / 55

108 Example Least squares problems Let A = ( ) 1 1 = I 0 0 ( ) ( ) 1 1 2, 1 1 Its pseudo inverse is A = 1 ( ) ( ) 1 1 1/ 2 0 I = ( ) Then A 2 = A and (A ) 2 = 1 2 A, i.e. (A 2 ) (A ) 2. TUHH Heinrich Voss Numerical Linear Algebra Chap. 4: Perturbation and Regularisation / 55

109 Least squares problems Perturbation of least squares problems Consider the linear least squares problem Ax b 2 = min! (1) with A R (m,n), rank(a) = r, and a perturbed problem A(x + x) (b + b) 2 = min!, (2) where we incorporate only perturbations of the right hand side b, but not of the system matrix A. TUHH Heinrich Voss Numerical Linear Algebra Chap. 4: Perturbation and Regularisation / 55

110 Least squares problems Perturbation of least squares problems Consider the linear least squares problem Ax b 2 = min! (1) with A R (m,n), rank(a) = r, and a perturbed problem A(x + x) (b + b) 2 = min!, (2) where we incorporate only perturbations of the right hand side b, but not of the system matrix A. Let x = A b and x + x = A (b + b) the pseudo normal solution of (1) and (2), respectively. TUHH Heinrich Voss Numerical Linear Algebra Chap. 4: Perturbation and Regularisation / 55

111 Least squares problems Perturbation of least squares problems Consider the linear least squares problem Ax b 2 = min! (1) with A R (m,n), rank(a) = r, and a perturbed problem A(x + x) (b + b) 2 = min!, (2) where we incorporate only perturbations of the right hand side b, but not of the system matrix A. Let x = A b and x + x = A (b + b) the pseudo normal solution of (1) and (2), respectively. Then x = A b, and from A 2 = 1 σ r it follows x 2 A 2 b 2 = 1 σ r b 2. TUHH Heinrich Voss Numerical Linear Algebra Chap. 4: Perturbation and Regularisation / 55

112 Least squares problems Perturbation of least squares problems It holds that r x 2 2 = c 2 i σ 2 i=1 i 1 σ 2 1 r ci 2 = 1 r 2 σ1 2 c i u i. i=1 i=1 2 TUHH Heinrich Voss Numerical Linear Algebra Chap. 4: Perturbation and Regularisation / 55

113 Least squares problems Perturbation of least squares problems It holds that Obviously, x 2 2 = r c 2 i σ 2 i=1 i 1 σ 2 1 r ci 2 = 1 r 2 σ1 2 c i u i. i=1 r c i u i is the projection of b to the range of A. Therefore it follows i=1 for the relative error i=1 x 2 σ 1 b 2. (3) x 2 σ r P range(a)b 2 2 TUHH Heinrich Voss Numerical Linear Algebra Chap. 4: Perturbation and Regularisation / 55

114 Least squares problems Perturbation of least squares problems It holds that Obviously, x 2 2 = r c 2 i σ 2 i=1 i 1 σ 2 1 r ci 2 = 1 r 2 σ1 2 c i u i. i=1 r c i u i is the projection of b to the range of A. Therefore it follows i=1 for the relative error i=1 x 2 σ 1 b 2. (3) x 2 σ r P range(a)b 2 2 This inequality specifies, how the relative error of the right hand side of a linear least squares problem effects the solution of the problem TUHH Heinrich Voss Numerical Linear Algebra Chap. 4: Perturbation and Regularisation / 55

115 condition Least squares problems Definition For A R (m,n) let A = UΣV T be the singular value decomposition, and let rank(a) = r. Then κ 2 (A) := σ 1 σ r is called condition of A. TUHH Heinrich Voss Numerical Linear Algebra Chap. 4: Perturbation and Regularisation / 55

116 condition Least squares problems Definition For A R (m,n) let A = UΣV T be the singular value decomposition, and let rank(a) = r. Then κ 2 (A) := σ 1 σ r is called condition of A. If A R (n,n) is nonsingular then this definition coincides with the one with respect to the Euclidean norm for quadratic matrices given before. TUHH Heinrich Voss Numerical Linear Algebra Chap. 4: Perturbation and Regularisation / 55

117 condition Least squares problems Definition For A R (m,n) let A = UΣV T be the singular value decomposition, and let rank(a) = r. Then κ 2 (A) := σ 1 σ r is called condition of A. If A R (n,n) is nonsingular then this definition coincides with the one with respect to the Euclidean norm for quadratic matrices given before. κ 2 (A T A) = κ 2 (A) 2 Hence, the normal equation of a linear least squares problem are much worse conditioned than the system matrix of the problem. TUHH Heinrich Voss Numerical Linear Algebra Chap. 4: Perturbation and Regularisation / 55

118 Least squares problems Perturbed least squares problems For perturbations of the system matrix the following theorem holds TUHH Heinrich Voss Numerical Linear Algebra Chap. 4: Perturbation and Regularisation / 55

119 Least squares problems Perturbed least squares problems For perturbations of the system matrix the following theorem holds Theorem Assume that A R (m,n), m n, is not deficient, i.e. rank(a) = n. Let x be the solution of the least squares problem (1) and x be the solution of the perturbed problem (A + A)x (b + b) 2 = min!, (4) where ( A 2 ε := max, b ) 2 < 1 A 2 b 2 κ 2 (A) = σ n(a) σ 1 (A). (5) TUHH Heinrich Voss Numerical Linear Algebra Chap. 4: Perturbation and Regularisation / 55

120 Least squares problems Perturbed least squares problems For perturbations of the system matrix the following theorem holds Theorem Assume that A R (m,n), m n, is not deficient, i.e. rank(a) = n. Let x be the solution of the least squares problem (1) and x be the solution of the perturbed problem (A + A)x (b + b) 2 = min!, (4) where ( A 2 ε := max, b ) 2 < 1 A 2 b 2 κ 2 (A) = σ n(a) σ 1 (A). (5) Then it holds that x x 2 x 2 ( ) 2κ2 (A) ε cos θ + tan θ κ2 2(A) + O(ε 2 ), (6) where θ is the angle between b and its projection to range(a). TUHH Heinrich Voss Numerical Linear Algebra Chap. 4: Perturbation and Regularisation / 55

121 Example Regularization Consider the orthogonal projection of a given function f : [0, 1] R to the space Π n 1 of polynomial of maximum degree n 1 with respect to the scalar product f, g := 1 0 f (x)g(x) dx. TUHH Heinrich Voss Numerical Linear Algebra Chap. 4: Perturbation and Regularisation / 55

122 Example Regularization Consider the orthogonal projection of a given function f : [0, 1] R to the space Π n 1 of polynomial of maximum degree n 1 with respect to the scalar product f, g := 1 0 f (x)g(x) dx. Choosing the basis {1, x,..., x n 1 } one obtains the linear system with Ay = b (1) A = (a ij ) i,j=1,...,n, a ij := 1 i + j 1, (2) the so called Hilbert Matrix, and b R n, b i := f, x i 1. TUHH Heinrich Voss Numerical Linear Algebra Chap. 4: Perturbation and Regularisation / 55

123 Example ct. Regularization For dimensions n = 10, n = 20 and n = 40 we choose the right hand side of (1) such that y = (1,..., 1) T is the unique solution, and we solve the resulting system by the known methods. TUHH Heinrich Voss Numerical Linear Algebra Chap. 4: Perturbation and Regularisation / 55

124 Example ct. Regularization For dimensions n = 10, n = 20 and n = 40 we choose the right hand side of (1) such that y = (1,..., 1) T is the unique solution, and we solve the resulting system by the known methods. The LU factorization with column pivoting (in MATLAB A\b), the Cholesky factorization, the QR decomposition of A and the singular value decomposition of A yield the following errors with respect to the Euclidean norm: TUHH Heinrich Voss Numerical Linear Algebra Chap. 4: Perturbation and Regularisation / 55

125 Example ct. Regularization For dimensions n = 10, n = 20 and n = 40 we choose the right hand side of (1) such that y = (1,..., 1) T is the unique solution, and we solve the resulting system by the known methods. The LU factorization with column pivoting (in MATLAB A\b), the Cholesky factorization, the QR decomposition of A and the singular value decomposition of A yield the following errors with respect to the Euclidean norm: n = 10 n = 20 n = 40 LU factorization 5.24 E E E+2 Cholesky 7.15 E-4 numer. nicht pos. def. QR decomposition 1.41 E E E+3 SVD 8.24 E E E+2 TUHH Heinrich Voss Numerical Linear Algebra Chap. 4: Perturbation and Regularisation / 55

126 Example ct. Regularization A similar behavior is observed for the least square problem. For n = 10, n = 20 and n = 40 and m = n + 10 we consider the least squares problem Ax b 2 = min! with the Hilbert matrix A R (m,n), where b is chosen such that x = (1,..., 1) T solves the problem with residual Ax b = 0. TUHH Heinrich Voss Numerical Linear Algebra Chap. 4: Perturbation and Regularisation / 55

127 Example ct. Regularization A similar behavior is observed for the least square problem. For n = 10, n = 20 and n = 40 and m = n + 10 we consider the least squares problem Ax b 2 = min! with the Hilbert matrix A R (m,n), where b is chosen such that x = (1,..., 1) T solves the problem with residual Ax b = 0. n = 10 n = 20 n = 40 Normal equations 2.91 E E E+2 QR factorization 1.93 E E E+1 SVD 4.67 E E E+2 TUHH Heinrich Voss Numerical Linear Algebra Chap. 4: Perturbation and Regularisation / 55

128 Regularization Regularization For badly conditioned least squares problem or linear systems the following approach can yield reliable solutions: TUHH Heinrich Voss Numerical Linear Algebra Chap. 4: Perturbation and Regularisation / 55

129 Regularization Regularization For badly conditioned least squares problem or linear systems the following approach can yield reliable solutions: Determine the singular value decomposition A = UΣVT of A, and define { Σ σ 1 τ = diag(η i δ ji ), η i := i falls σ i τ, 0 sonst, where τ > 0 is a given threshold, and A τ := V Σ τ U T, x τ := A τ b. TUHH Heinrich Voss Numerical Linear Algebra Chap. 4: Perturbation and Regularisation / 55

130 Regularization Regularization For badly conditioned least squares problem or linear systems the following approach can yield reliable solutions: Determine the singular value decomposition A = UΣVT of A, and define { Σ σ 1 τ = diag(η i δ ji ), η i := i falls σ i τ, 0 sonst, where τ > 0 is a given threshold, and A τ := V Σ τ U T, x τ := A τ b. A τ is called effektive pseudo inverse of A. This method of approximatively solving Ax = b is called regularization by truncation TUHH Heinrich Voss Numerical Linear Algebra Chap. 4: Perturbation and Regularisation / 55

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