EECS 275 Matrix Computation

Transcription

1 EECS 275 Matrix Computation Ming-Hsuan Yang Electrical Engineering and Computer Science University of California at Merced Merced, CA Lecture 12 1 / 18

2 Overview QR decomposition by Householder transformation QR decomposition by Givens rotation 2 / 18

3 Reading Chapter 10 of Numerical Linear Algebra by Llyod Trefethen and David Bau Chapter 5 of Matrix Computations by Gene Golub and Charles Van Loan Chapter 5 of Matrix Analysis and Applied Linear Algebra by Carl Meyer 3 / 18

4 Householder triangularization In Gram-Schmidt process A R 1 R 2 R }{{ n = Q } R 1 has orthonormal columns. The product R = Rn 1 triangular R 1 2 R 1 1 is upper In Householder triangularization, a series of elementary orthogonal matrices Q k is applied to the left of A Q n Q 2 Q }{{} 1 A = R Q is upper triangular. The product Q = Q1 Q 2 Q n and therefore A = QR is a QR factorization of A is orthogonal, 4 / 18

5 Geometry of elementary projectors For u, x IR n, s.t. u = 1 Orthogonal projectors onto span{u} and u are P u = uu u u, and P u = I uu u u For u 0, the Householder transformation or the elementary reflector about u is R = I 2 uu u u or R = I 2uu when u = 1, and R = R = R 1 5 / 18

6 Triangularization by introducing zeros The matrix Q k is chosen to introduce zeros below the diagonal in the k-th column while preserving all the zeros previously introduced Q Q 2 0 Q A Q 1 A Q 2 Q 1 A Q 3 Q 2 Q 1 A Q k operates on row k,..., m (changed entries are denoted by boldface or and blank entries are zero) At beginning of step k, there is a block of zeros in the first k 1 columns of these rows The application of Q k forms linear combinations of these rows, and the linear combination of the zero entries remain zero After n steps, all the entries below the diagonal have been eliminated and Q n Q 2 Q 1 A = R is upper triangular 6 / 18

7 Householder reflectors At beginning of step k, there is a block of zeros in the first k 1 columns of these rows Each Q k is chosen to be [ ] I 0 Q k = 0 F where I is the (k 1) (k 1) identity matrix and F is an (m k + 1) (m k + 1) orthogonal matrix Multiplication by F has to introduce zeros into the k-th column The Householder algorithm chooses F to be a particular matrix called Householder reflector At step k, the entries k,..., m of the k-th column are given by vector x IR m k+1 7 / 18

8 Householder transformation (cont d) To introduce zeros into k-th column (x IR m k+1 ), the Householder transformation F should x F 0 x = F x =.. = x e 1 = αe 1 0 The reflector F will reflect the space IR m k+1 across the hyperplane H orthogonal to u = x e 1 x A hyperplane is characterized by a vector u = x e 1 x 8 / 18

9 Householder transformation (cont d) Every point x IR m is mapped to a mirror point and hence F x = (I 2 uu u u )x = x 2u( u x u u ) F = (I 2 uu u u ) Will fix the +/- sign in the next slide 9 / 18

10 The better of two Householder reflectors Two Householder reflectors (transformations) For numerical stability pick the one that moves reflect x to the vector x e 1 that is not to close to x itself, i.e., x e 1 x in this case In other words, the better of the two reflectors is u = sign(x 1 ) x e 1 + x where x 1 is the first element of x (sign(x 1 ) = 1 if x 1 = 0) 10 / 18

11 Householder QR factorization Algorithm: for k = 1 to n do x = A k:m,k u k = sign(x 1 ) x 2 e 1 + x u k = u k u k 2 A k:m,k:n = (I 2u k u k )A k:m,k:n end for Recall Q k = [ ] I 0 0 F Upon completion, A has been reduced to upper triangular form, i.e., R in A = QR Q = Q n Q 2 Q 1 or Q = Q 1 Q 2 Q n 11 / 18

12 QR decomposition with Householder transformation Want to compute QR decomposition A with Householder transformation A = Need to find a reflector for first column of A, x = [12, 6, 4] to x e 1 = [14, 0, 0] u= x e 1 x = [2, 6, 4] = 2[1, 3, 2] 6/7 3/7 2/ F 1 =I 2 uu u u =, F 1 A = 3/7 2/7 6/7 2/7 6/7 3/7 Next need to zero out A 32 and apply the same process to [ ] A = / 18

13 QR decomposition with Householder (cont d) With the same process F 2 = 0 7/25 24/ /25 7/25 Thus, we have 6/7 69/175 58/175 Q = Q 1 Q 2 = 3/7 158/175 6/175 2/7 6/35 33/ R = Q 2 Q 1 A = Q A = The matrix Q is orthogonal and R is upper triangular 13 / 18

14 Givens rotations Givens rotation: orthogonal transform to zero out elements selectively c s 0 i G(i, k, θ) = s c 0 k i k where c = cos(θ) and s = sin(θ) for some θ Pre-multiply G(i, k, θ) amounts to a counterclockwise rotation θ in the (i, k) coordinate plane, y = G(i, k, θ)x cx i sx k j = i y j = sx i + cx k j = k j i, k x j 14 / 18

15 Givens rotations (cont d) Can zero out y k = sx i + cx k = 0 by setting c = x 2 i x i + x 2 k, s = x k, θ = arctan(x k /x i ) xi 2 + xk 2 QR decomposition can be computed by a series of Givens rotations Each rotation zeros an element in the subdiagonal of the matrix, forming R matrix, Q = G 1... G n forms the orthogonal Q matrix Useful for zero out few elements off diagonal (e.g., sparse matrix) Example A = Want to zero out A 31 = 4 with rotation vector (6, 4) to point along the x-axis, i.e., θ = arctan( 4/6) 15 / 18

16 QR factorization with Givens rotation With θ we have the orthogonal Givens rotation G G 1 = 0 cos(θ) sin(θ) = sin(θ) cos(θ) Pre-multiply A with G G 1 A = Continue to zero out A 21 and A 32 and form a triangular matrix R The orthogonal matrix Q = G 3 G 2 G 1, and G 3 G 2 G 1 A = Q A = R for QR decomposition 16 / 18

17 Gram-Schmidt, Householder and Givens Householder QR is numerically more stable Gram-Schmidt computes orthonormal basis incrementally Givens rotation is more useful for zero out few selective elements 17 / 18

18 Eigendecomposition Also known as spectral decomposition A is a square matrix A = QDQ 1 where Q is a square matrix whose columns are eigenvector and D is a diagonal matrix whose elements are the corresponding eigenvalues With eigendecomposition AQ = QD Aq i = λ i q i where λ i and q i are eigenvalues and eigenvectors of Ax = λx The eigenvectors are usually normalized but not necessarily If A can be eigendecomposed with all non-zero eigenvalues A 1 = QD 1 Q 1 18 / 18