Conjugate Gradient (CG) Method

Transcription

1 Conjugate Gradient (CG) Method by K. Ozawa 1 Introduction In the series of this lecture, I will introduce the conjugate gradient method, which solves efficiently large scale sparse linear simultaneous equations. 2 Minimization problem Consider the quadratic function given by Q (x) = 1 2 xt A x b T x, A R n n, b, x R n, (1) where A is a symmetric positive definite matrix. This quadratic function attains its minimum at the solution of the linear algebraic equation A x = b, which will be denoted by x. A variety of methods for solving this minimization problem have been derived. The general form of these algorithms is x k+1 = x k α k p k, k = 0, 1,..., (2) where p k is the direction vector at the kth iteration. The parameter α k is determined so as to minimize the objective function Q(x), that is, α k is the value satisfying Since in the expression Q (x k α k p k ) = min α Q (x k α p k ). (3) Q (x k α k p k ) = 1 2 (pt k A p k ) α 2 k p T k (A x k b) α k xt k (Ax k 2 b), (4) the coefficients of α 2 k is positive, this function attains its minimum at α k = p T k (A x k b)/(pt k A p k ), (5) for the fixed x k and p k. Next we consider the method for choosing the direction vectors p k. 3 Univariate iteration If we put p k = e i (the unit vector whose ith entry is and the others are 0) in (2), then with the α k given by (5) we have ( ) x k+1 = x k α k e i = x k 1 a a ij x k j b i e i, (6) ii j=1 1

2 since e T i A e i = a ii and e T i (A x b) = a ij x j b i. This means that the point x k+1 given by (6) is the minimum point which can be obtained by changing only the ith component of x k. Thus, the first n successive iterations of (6) with j=1 p 0 = e 1, p 1 = e 2,..., p n 1 = e n (7) is equivalent to the one iteration of the well-known Gauss Seidel method. 4 Steepest descent method On the other hand, if we choose p k as the gradient vector of Q(x) at the point x k, then we have the iteration method given by since where ( Q Q(x) =,..., Q ) T. x 1 x n This method is called the steepest descent method. 5 Conjugate direction method Let us consider the case that p k are chosen as x k+1 = x k α k (A x k b), (8) p k = Q(x k ) = A x k b, (9) p T i A p j = 0, i j. (10) When the direction vectors p k are taken in this way, the method is called the conjugate direction method. The vectors p i and p j satisfying (10) are said to be A-conjugate. The conjugate direction method converges in n steps, if no roundoff occurs. Here we show the convergence of the method. First of all, notice that (A x k+1 b) T p j = (A x k α k A p k b) T p j = (A x k b) T p j α k (A p k ) T p j = (A x k b) T p j, and as a result, we have (A x k+1 b) T p j = { (A x k b) T p j, j < k, 0, j = k. (11) 2

3 Thus we have (A x n b) T p j = (A x n 1 b) T p j = = (A x j+1 b) T p j = 0, j = 0,...,n 1. Therefore, the vector A x n b is orthogonal to the n linearly independent vectors p 0,...,p n 1, that is, A x n b = 0. Here we consider the method of generating the conjugate vectors p j. One might think of the method of selecting eigenvectors of A as p j (j = 1,..., n), since the two distinct eigenvectors of A, say p i, p j, satisfy p T i A p j = λ j pt i p j = 0. However, in general, the computation of all eigenvectors is far more expensive than solving linear equations, so that this method is not practical at all. The conjugate gradient method to be explained in the next section generates the conjugate direction vectors by using a relatively cheap procedure. 6 Conjugate gradient method Here we show the algorithm: CG-1 1: Choose a small value ε > 0 and an initial guess x 0 2: p 0 = r 0 = b A x 0, and compute (r 0, r 0 ) 3: k = 0 4: while r k / b ε do 5: α k = (r k, p k )/(p k, A p k ) 6: x k+1 = x k α k p k 7: r k+1 = r k + α k A p k 8: β k = (p k, A r k+1 )/(p k, A p k ) 9: p k+1 = r k+1 + β k p k 10: k = k : end while The vectors p k generated by the algorithm satisfy the conjugacy condition (p k, A p j ) = 0, 0 j < k, k = 1,..., n 1, (12) and the residuals r j (= b A x j ) satisfy the orthogonality condition(see Appendix) (r k, r j ) = 0, 0 j < k, k = 1,...,n 1. (13) Moreover, from the definitions of α j and β j we have (p j, r j+1 ) = (p j, r j ) + α j (p j, A p j ) = 0, j = 0, 1,... (14) (p j, A p j+1 ) = (p j, A r j+1 ) + β j (p j, A p j ) = 0, j = 0, 1,... (15) Using the relations above, we have a variant of CG-1: 3

4 CG-2 1: Choose a small value ε > 0 and an initial guess x 0 2: p 0 = r 0 = b A x 0, and compute (r 0, r 0 ) 3: k = 0 4: while r k / b ε do 5: α k = (r k, r k )/(p k, A p k ) 6: x k+1 = x k α k p k 7: r k+1 = r k + α k A p k 8: β k = (r k+1, r k+1 )/(r k, r k ) 9: p k+1 = r k+1 + β k p k 10: k = k : end while The number of the operations to be performed per step in this algorithm is shown in Table 1, which is a tremendous improvement over CG-1. Table 1.Number of the operations per step in CG-2. matrix-vector product 1 inner product 2 vector addition 3 scalar division 1 scalar comparison 1 4

5 Here we show the experimantal result for some dimensional equation ( xk x k log ) 10 x ( ) rk log 10 b k Relative residual and error versus iteration number k. References [1] James M. Ortega, Introduction to Parallel and Vector Solution of Linear Systems, 1989, Prenum Press, New York. [2] Gilbert W. Stewart, Afternotes Goes to Graduate School, 1997, SIAM, Philadelphia. 5

6 Problem 1. For the function P(x) given by P(x) = 1 2 (x x ) T A (x x ), show that the relation P(x) = Q(x) (x ) T Ax, where x is the solution of A x = b, and Q(x) is the function given by (1). 2. Prove equation (4). 3. Show that the equation (9). 4. Show that the vectors p 0,...,p n 1 given by (10) are linearly independent. 5. Show that equation (11) is valid. 6. Show that diagonal elements of symmetric positive definite matrices are positive. 7. For CG-1 make the table as Table Show that the two algorithms, CG-1 and CG-2, are equivalent. 9. Show that the matrix given by A = is symmetric positive definite Solve the equation A x = b by CG-1 and CG-2, and compare the CPU times for the case that A is the matrix given by above and the vector b is given by b = A u, where the dimension of the equation is n = and each component of u is a random number in the range (0, 1). 6

7 Appendix 1 Inner product Let x = (x 1,...,x n ) T and y = (y 1,...,y n ) T be real vectors in R n. Then we define the inner product of x and y by (x, y) = x T y = x i y i. (16) The inner product has the following properties: i=1 (x, y) = (y, x) (α x, y) = α (x, y), α is scalar (x, y + z) = (x, y) + (x, z) (x, x) 0, = holds, only if x = 0. (17) For nonzero vectors x, y R n, if (x, y) = 0 then the vectors x and y are said to be orthogonal. If the vector x is orthogonal with the linearly independent n vectors p i (i = 1,...,n) in R n, then x = 0, since from the assumption we have p T 1 p T 2. x = 0, p T n and the matrix in the left-hand side is nonsingular, so that x = 0. Next we show the set of n vectors p i (i = 1,..., n) which are orthogonal with each other are linearly independent. Since, if not so, that is, if p i (i = 1,...,n) are linearly dependent, then there exist constants c i (i = 1,...,n), not necessarily all zero, such that From this we have for all j c 1 p c n p n = 0. 0 = (c 1 p c n p n, p j ) = c j (p j, p j ), which means c j = 0. This contradicts the assumption. 2 Eigenvalues and eigenvectors of real symmetric matrices When a real matrix A satisfies a ij = a ji, that is, A T = A, the matrix A is called real symmetric matrix. The eigenvalues and eigenvectors of real symmetric matrices have the following properties: 7

8 1. Eigenvalues are real Let λ and x be any eigenvalue end eigenvector of A, respectively. Then we have (A x, x) = (λx, x) = λ( x, x), where denotes complex conjugate. On the other hand, from the property of inner product we have (A x, x) = ( x, Ax) = λ ( x, x). These two expressions mean λ = λ, since ( x, x) Orthogonality of eigenvectors Let λ i and λ j be the eigenvalues of A and assume λ i λ j. Then we have and since A is symmetric, which implies Thus we have (x i, x j ) = 0. (Ax i, x j ) = λ i (x i, x j ), (Ax i, x j ) = (x i, Ax j ) = λ j (x i, x j ), (λ i λ j ) (x i, x j ) = 0. 3 Quadratic form For a real symmetric matrix A = (a ij ) and a real vector x = (x 1,...,x n ) T, the quantity given by Q (x) = x T A x = (x, A x) = a ij x i x j (18) is called quadratic form. For any x 0, if Q (x) > 0 ( 0), then the matrix A is said to be positive (semi-)definite. Positive definite matrices have the following properties: 1. The diagonal elements of symmetric positive definite matrix is positive. This is clear from a ii = (e i, A e i ). 2. All the eigenvalues of symmetric positive definite matrix A are positive. For any x 0, if we transform x by y = T x, where T = (u 1,...,u n ), and u i is the eigenvector of A corresponding to λ i, then i=1 j=1 which means λ i > 0 for all i. 0 < x T A x = y T (T T A T) y = y T (T 1 A T) y = λ i yi 2, i=1 8

9 4 Theorem Here we show again the algorithm CG-1: 1: Choose a small value ε > 0 and initial guess x 0 2: p 0 = r 0 = b A x 0, and compute (r 0, r 0 ) 3: k = 0 4: while (r k, r k ) ε do 5: α k = (r k, p k )/(p k, A p k ) 6: x k+1 = x k α k p k 7: r k+1 = r k + α k A p k 8: β k = (p k, A r k+1 )/(p k, A p k ) 9: p k+1 = r k+1 + β k p k 10: k = k : end while Theorem Let A be an n n symmetric positive definite matrix, and x be the solution of the equation A x = b. Then the vectors p k generated by CG-1 algorithm satisfy and p k 0 unless x k = x. p T k A p j = 0, 0 j < k, k = 1,...,n 1, (19) r T k r j = 0, 0 j < k, k = 1,...,n 1, (20) Proof By the definitions of α j, β j and those of p j, r j+1, we have p T j r j+1 = pt j r j + α j p T j A p j = 0, j = 0, 1,..., (21) p T j A p j+1 = p T j A r j+1 + β j p T j A p j = 0, j = 0, 1,... (22) Here we assume, as an induction hypothesis, that (19) and (20) hold for some k < n 1. Then we must show these hold for k +1. Since p 0 = r 0, these hold for k = 1. For any j < k, using the 7th and 9th lines in CG-1, we have r T j r k+1 = rt j (r k + α k A p k ) = r T j r k + α k p T k A r j = r T j r k + α k p T k A (p j β j 1 p j 1 ) = 0, (23) since all three terms in the last line vanish by induction hypothesis. Moreover, we have from (21) and (22) r T k r k+1 = (p k β k 1 p k 1 ) T r k+1 = β k 1 p T k 1 r k+1 = β k 1 p T k 1 (r k + α k A p k ) = 0. 9

10 Thus we have shown that (20) is true also for k + 1. Next we have for any j < k p T j A p k+1 = pt j A (r k+1 + β k p k ) = pt j A r k+1 = α 1 j (r j+1 r j ) T r k+1 = 0, (24) if α j 0, which we will show later. Therefore (19) is true also for k + 1. Finally, we show α j 0. By the definition of p j and (22) we have Hence we have r T j p j = rt j (r j + β j 1 p j 1 ) = rt j r j. α j = r T j r j /p T j A p j. Therefore, if α j = 0, then r j = 0, that is, x j = x so that the process stops with x j. 5 Differentiation by vectors Here we define the differentiation of the scalar-valued function J(x) with respect to its vector argument by J (x) := J x 1 J x 2. J x n According to this definition, we have ( ) 1 Q (x) = 2 xt A x b T x = A x b,. (25) (26) since and Q(x) = 1 2 b T x = a ii x 2 i + a ij x i x j, i j i=1 b i x i, i=1 x k Q(x) = a kk x k + j k x k (b T x) = b k. a kj x j = a kj x j, j=1 10