ACM106a - Homework 2 Solutions

Transcription

1 ACM06a - Homework 2 Solutions prepared by Svitlana Vyetrenko October 7, Chapter 2, problem 2.2 (solution adapted from Golub, Van Loan, pp.52-54): For the proof we will use the fact that if A C m m has an LU-factorization A = LU and has an upper bandwidth q and a lower bandwidth p, then U has an upper bandwidth q and L has a lower bandwidth p (see Golub, Van Loan, p.52, theorem 4..). Now let P A = LU be the factorization computed by Gaussian elimination with partial pivoting. P = P n... P - a product of permutation matrices. P T = [e s... e sn ], where s, s 2... s n is a permutation of, 2,... n. If s i > i + p, then it follows that the leading i i principal submatrix of P A is singular, since (P A) ij = a si,j for j =... s i p and s i p i. This implies that U and A are singular, thus, we reach a contradiction. Therefore, s i i + p for i =... n, thus, P A has an upper bandwidth p+p = 2p. Thus, by the above mentioned fact we see that U has an upper bandwidth 2p. Note that pivoting destroys the band structure in a sense that U ends up having a higher bandwidth, while nothing at all can be said about the band structure of L (in fact, since L = P (L n P n... L P ) the only thing we can say is that L contains at most p + nonzero elements per column). 2. Chapter 2, problem 2.6: For the first step of Gaussian elimination with partial pivoting, the entry of maximum modulus is a, therefore, no row interchange is necessary. a a 2... a n Let A () be the matrix obtained after the first step: A () = 0 a () a () 2n a () n2... a () nn Let s show that A () is again column diagonally dominant: a () ij = n a ij a ja i a a ij + a ja i a < ( a jj a j ) + a j ( a a j ) a = a jj a ja j a = a () jj, where the second inequality and the fact that a jj a > a j a j follow from the diagonal dominance of A.

2 Thus, the pivot for the second step is a () 22 and, consequently, no row interchange is necessary. Similarly, this process can be continued to show that each step of Gaussian elimination results in a strictly column diagonally dominant matrix, therefore, the element with the maximum absolute value will be on the diagonal and no row interchanges will take place.. Chapter 2, problem 2.: We want to see how long it takes to solve Ax = b (i.e. which solver is being used to solve the system) depending on the structure of A (for example, if A is symmetric, symmetric positive definite, triangular, etc.) You can find a good description of what backslash actually does depending on the structure of A at (a) Note that A is symmetric positive definite matrix (as a remark, A is invertible with probability ). We expect that the system is going to be solved using Cholesky factorization (the flop count is m ). (b) Run it just to make sure that no operating system effects are incorporated into the estimated time for part (a). (c) A 2 is not symmetric. We expect that it is going to be solved using LU-factorization. Our guess is confirmed by time estimate (recall that Gaussian elimination with partial pivoting takes 2m flops), which is twice as large as the time count in part (a). (d) A is symmetric positive definite (note that if λ is the eigenvalue of B, then λ σ is the eigenvalue of B σi). Therefore, Cholesky factorization is used to solve the system. It costs roughly m flops, which agrees with part (a). (e) A 4 is still a symmetric matrix. However, it is not positive definite since it has at least one eigenvalue which is less than zero. According to the mldivide algorithm documentation available at we can assume that because of the symmetry of A 4 MATLAB attempts to run Cholesky factorization first, which subsequently fails and then proceeds with the indefinite symmetric factorization. (f) A 5 is upper triangular, thus the system is solved using back substitution - roughly m 2 flops. (g) A 6 is not symmetric again. The system is solved using Gaussian elimination with partial pivoting, which results in 2m flops. 4. Chapter 2, problem 2.2: (a) Given n distinct points x, x 2... x n and n corresponding function values f(x ), f(x 2 )... f(x n ) there exists a unique polynomial of degree n P n (x) such that P n (x i ) = f(x i ) (it is called Lagrange interpolating polynomial). Its explicit representation is: P n (x) = f(x i ). n x x j x i x j (b) Therefore, the entries of A are: function pr22b; for n=2:0 m=2*n-; n j=,j i y k x j x i x j, k =... m, i =... n. i= j=,j i 2

3 f=:n; x=-+2*(f-)/(n-); f=:m; y=-+2*(f-)/(m-); A=ones(m,n); for i=:m for j=:n for f=:n if (f~=j) A(i,j)=A(i,j)*(y(i)-x(f))/(x(j)-x(f)); array(n,)=n; array(n,2)=log(norm(a,inf)); array(n,)=n*log(2)-log(exp()*(n-)*log(n)); plot(array(2:0,), array(2:0,2), array(2:0,), array(2:0,)); 6 4 log(norm(a,inf)) as computed suggested asymptotic formula 2 log(norm(a,inf)) n (c) Note that for our problem (2.6) is equivalent to (2.9). function pr22c; for n=2:0 m=2*n-; f=:n; x=-+2*(f-)/(n-); f=:m; y=-+2*(f-)/(m-); val=ones(n,); A=ones(m,n);

4 for i=:m for j=:n for f=:n if (f~=j) A(i,j)=A(i,j)*(y(i)-x(f))/(x(j)-x(f)); array(n,)=n; array(n,2)=(norm(a,inf)/norm(a*val, inf)); plot(array(2:0,),log( array(2:0,2))); 5 0 log K n (d) Now let s use the code of the previous problem to sketch the computed polynomial interpolation for n = and n = 0: 4

5 n=, K=24.66 p(y) y n=0, K=2.084e+06 p(y) y 5. Problem 5: (a) We know that (I + uv T )x = b, i.e. x + uv T x = b. Multiply the last identity by v T : v T x = vt b + v T u, therefore thus x + x = (I uvt b + v T u = b, uvt + v T u )b 5

6 and if v T u. (I + uv T ) = I uvt + v T u Note that when v T u = we have (I + uv T )u = u + u(v T u) = u u = 0, hence 0 is the eigenvalue of I + uv T, therefore, I + uv T is singular. (b) Using the result of part (a) we have (A + uv T ) = (I + A uv T ) A = (I + v T A u A uv T )A = A + v T A u A uv T A. (c) Having a fast solver for Ax = b means that we can easily compute x = A b. If Hx = b, we have x = H b = (A + uv T )b = A b + v T A u A uv T A b. Now since we know how to compute A b and A u, we can obtain the solution to the system. (d) Note that if A is orthonormal, A = A T and therefore x = A T b function x=problem5(a,u,v,b) pr=a *u; pr2=a *b; x=pr2-/(+v *pr)*pr*v *pr2; + v T A T u AT uv T A T b. 6