lecture 3 and 4: algorithms for linear algebra

Transcription

1 lecture 3 and 4: algorithms for linear algebra STAT 545: Introduction to computational statistics Vinayak Rao Department of Statistics, Purdue University August 30, 2016

2 Solving a system of linear equations Consider AX = b, where A is N N, and X and b are N k. Solve for X: X = A 1 b Calculate the inverse of A and multiply? No! Directly solving for X is faster, and more stable numerically A 1 need not even exist > solve(a,b) # Directly solve for b > solve(a) %*% b # Return inverse and multiply dont-invert-that-matrix/ 1/23

3 Gauss-Jordan elimination A X = b A [ X, A 1 ] = [ b, I ] x 1 v 11 v 12 v x 2 v 21 v 22 v 23 = x 3 v 31 v 32 v Manipulate to get: I [ X, A 1 ] = [ ĉ 1, Ĉ 2 ] At step i: Make element a ii = 1 (by scaling or pivoting) Set other elements in column i to 0 by multiplying and subtracting that row 2/23

4 Gauss-Jordan elimination x 1 v 11 v 12 v x 2 v 21 v 22 v 23 = x 3 v 31 v 32 v /23

5 Gauss-Jordan elimination x 1 v 11 v 12 v x 2 v 21 v 22 v 23 = x 3 v 31 v 32 v Multiply row 1 by 2 and subtract 3/23

6 Gauss-Jordan elimination x 1 v 11 v 12 v x 2 v 21 v 22 v 23 = x 3 v 31 v 32 v /23

7 Gauss-Jordan elimination x 1 v 11 v 12 v x 2 v 21 v 22 v 23 = x 3 v 31 v 32 v Subtract row 1 3/23

8 Gauss-Jordan elimination x 1 v 11 v 12 v x 2 v 21 v 22 v 23 = x 3 v 31 v 32 v Pivot 3/23

9 Gauss-Jordan elimination x 1 v 11 v 12 v x 2 v 21 v 22 v 23 = x 3 v 31 v 32 v Continue till we get an identity matrix 3/23

10 Gauss-Jordan elimination x 1 v 11 v 12 v x 2 v 21 v 22 v 23 = x 3 v 31 v 32 v What is the cost of this algorithm? 3/23

11 Gauss elimination with back-substitution A [ X, A 1 ] = [ b, I ] O(N 3 ) manipulation to get: U [ X, A 1 ] = [ ĉ 1, Ĉ 2 ] Here, U is an upper-triangular matrix. Cannot just read off solution. Need to backsolve. 4/23

12 LU decomposition What are we actually doing? A = LU Here L and U are lower and upper triangular matrices. Is this always possible? u 11 u 12 u 13 L = l , U = 0 u 22 u 23 l 31 l u 33 [ ] 0 1 A = 1 0 PA = LU, P is a permutation matrix Crout s algorithm, O(N 3 ), stable, L, U can be computed in place. 5/23

13 Backsubstitution AX = b LUX = Pb First solve Y by forward substitution LY = Pb y 1 ˆb 1 l y 2 = ˆb 2 l 31 l 32 1 y 3 ˆb 3 Then solve X by back substitution UX = Y 6/23

14 Comments LU-decomposition can be reused for different b s. Calculating LU decomposition: O(N 3 ). Given LU decomposition, solving for X: O(N 2 ). A = P 1 LU = ( 1) S N i=1 u ii (S: num. of exchanges) LUA 1 = PI, can solve for A 1. (back to Gauss-Jordan) 7/23

15 Cholesky decomposition If A is symmetric positive-definite: A = LL T Square-root of A More stable. Twice as efficient. Related: A = LDL T. 8/23

16 Eigenvalue decomposition An N N matrix A: a map from R N R N. An eigenvector v undergoes no rotation: Av = λv λ is the corresponding eigenvalue, and gives the rescaling. Let Λ = diag(λ 1,, λ N ) with λ 1 λ 2,..., λ N, and V = [v 1,, v N ] be the matrix of corresponding eigenvectors Real Symmetric matrices have real eigenvalues AV = VΛ different eigenvalues have orthogonal eigenvectors 9/23

17 Gauss-Jordan elimination x 1 x 2 x 3 x 4 32 = What is the solution? How about for b = [32.1, 22.9, 33.1, 30.9] T? Why the difference? the determinant? the inverse? the condition number? An ill-conditioned problem can strongly amplify errors. Even without any rounding error 10/23

18 Stability analysis The norm of a matrix A is A 2 = A = max v =1 Av For a symmetric, real matrix, A = λ max (A) (why?) For a general, real matrix, A = λ max (A T A) (why?) For A in R N N and any v R N, Av A v (why?) If A = BC, and all matrices are in R N N, A B C (why?) 11/23

19 Stability analysis For a perturbation δb let δx be the change in solution to Ax = b A(x + δx) = b + δb δx x δb is the relative change in the solution from the change b From b = Ax and δx = A 1 δb, we have: δx x A A 1 δb b Condition number of a matrix A is given by κ(a) = A A 1 12/23

20 Stability analysis κ(a) 1 (why?) For a real symmetric matrix, κ(a) = λ max(a) λ min (A) (why?) For a real matrix, κ(a) = λmax (A T A) λ min (A T A) (why?) Condition number is a property of a problem Stability is a property of an algorithm A bad algorithm can mess up a simple problem 13/23

21 Gaussian elimination with partial pivoting Consider reducing to upper triangular v 11 v 12 v 13 v 21 v 22 v 23 v 31 v 32 v 33 Gaussian elimination: divide row 1 by v 11 Partial pivoting: Pivot rows to bring max v 1 to top Can dramatically improve performance. E.g. [ ] 1e Why does it work? 14/23

22 Gaussian elimination with partial pivoting Recall Gaussian elimination decomposes A = LU and solves two intermediate problems. What are the condition numbers of L and U? Try [ ] 1e /23

23 QR decomposition In general, for A = BC, κ(a) κ(b)κ(c) (why?) QR decomposition: A = QR Here, R is an upper (right) triangular matrix. Q is an orthonormal matrix: Q T Q = I κ(a) = κ(q)κ(r) Can use to solve Ax = b (How?) Most stable decomposition Does this mean we should use QR decomposition? 16/23

24 Gram-Schmidt orthonormalization Given N vectors x 1,..., x N construct an orthonormal basis: u 1 = x 1 / x 1 ũ i = x i i 1 j=1 (xt i u j)u i, u i = ũ i / ũ i i = 2..., N Modified Gram-Schmidt u 1 = x 1 / x 1 ũ i = x i (x T i u 1)u 1, ũ i = x i (u T i u 1)u 2, u i = ũ i / ũ i 17/23

25 QR decomposition A = Q R a 11 a 12 a 13 q 11 q 12 q 13 r 11 r 12 r 13 a 21 a 22 a 23 = q 21 q 22 q 23 0 r 22 r 23 a 31 a 32 a 33 q 31 q 32 q r 33 QR decomposition: Gram-Schmidt on columns of A (can you see why?) Of course, there are more stable/efficient ways of doing this (Householder rotation/givens rotation) O(N 3 ) algorithms (though about twice as slow as LU) 18/23

26 Calculating eigenvalues and -vectors Let A be any real symmetric matrix. How does one calculate its largest eigenvalue and vector? Start with a random vector u 0 Define u 1 = Au 0, and normalize length. Repeat: u i = Au i 1, u i = u i / u i u i v 1 (Why?) The power method (Google s PageRank). How would we calculate λ 1? What if we wanted the second eigenvector? 19/23

27 QR algorithm Algorithm to calculate all eigenvalues/eigenvectors of a (not too-large) matrix Start with A 0 = A At iteration i: A i = Q i R i A i+1 = R i Q i Can be made this more stable/efficient. One of Dongarra & Sullivan (2000) s list of top 10 algoirithms. See also number 4, decompositional approach to matrix computations 20/23

28 Multivariate normal log(p(x µ, Σ)) = 1 2 (X µ)t Σ 1 (X µ) N 2 log 2π 1 log Σ 2 Σ = LL T Y = L 1 (X µ) (Forward solve) log(p(x µ, Σ)) = 1 2 YT Y N 2 log 2π log Σ Can also just forward solve for L 1 : LL 1 = I (Inverted triangular matrix isn t too bad) 21/23

29 Sampling a multivariate normal Sampling a univariate normal: Inversion method (default for rnorm?). Box-Muller transform: (Z 1, Z 2 ) : independent standard normals. Let Z N (0, I) X = µ + LZ Z = N (µ, L T L) 22/23

30 x y [ ] 1 0 Σ = /23

31 x y [ ] 4 0 Σ = /23

32 x y [ ] 4 2 Σ = /23

33 x y [ ] Σ = /23