Linear Least Squares. Using SVD Decomposition.

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1 Linear Least Squares. Using SVD Decomposition. Dmitriy Leykekhman Spring 2011 Goals SVD-decomposition. Solving LLS with SVD-decomposition. D. Leykekhman Linear Least Squares 1

2 SVD Decomposition. For any matrix A R m n there exist orthogonal matrices U R m m, V R n n and a diagonal matrix Σ R m n, i.e., σ σr Σ = 0 for m n with diagonal entries such that A = UΣV T σ 1 σ r > σ r+1 = = σ min {m,n} = 0 D. Leykekhman Linear Least Squares 2

3 SVD Decomposition. For any matrix A R m n there exist orthogonal matrices U R m m, V R n n and a diagonal matrix Σ R m n, i.e., σ 1... σ r 0 Σ =... for m n with diagonal entries such that A = UΣV T σ 1 σ r > σ r+1 = = σ min {m,n} = 0 D. Leykekhman Linear Least Squares 2

4 SVD Decomposition. The decomposition A = UΣV T is called Singular Value Decomposition (SVD). It is very important decomposition of a matrix and tells us a lot about its structure. It can be computed using the Matlab command svd. The diagonal entries of Σ are called the singular values of A. The columns of U are called left singular vectors and the columns of V are called right singular vectors. D. Leykekhman Linear Least Squares 3

5 SVD Decomposition. The decomposition A = UΣV T is called Singular Value Decomposition (SVD). It is very important decomposition of a matrix and tells us a lot about its structure. It can be computed using the Matlab command svd. The diagonal entries of Σ are called the singular values of A. The columns of U are called left singular vectors and the columns of V are called right singular vectors. Using the orthogonality of V we can write it in the form AV = UΣ We can interpret it as follows: there exists a special orthonormal set of vectors (i.e. the columns of V ), that is mapped by the matrix A into an orthonormal set of vectors (i.e. the columns of U). D. Leykekhman Linear Least Squares 3

6 Applications of SVD Decomposition. Given the SVD-Decomposition of A, A = UΣV T with σ 1 σ r > σ r+1 = = σ min {m,n} = 0 one may conclude the following: rank(a) = r, R(A) = R([u 1,..., u r ]), N(A) = R([v r+1,..., v n ]), R(A T ) = R([v 1,..., v r ]), N(A T ) = R([u r+1,..., u m ]). D. Leykekhman Linear Least Squares 4

7 Applications of SVD Decomposition. Moreover if we denote U r = [u 1,..., u r ], Σ r = diag(σ 1,..., σ r ), V r = [v 1,..., v r ], then we have A = U r Σ r V T r = u i vi T This is called the dyadic decomposition of A, decomposes the matrix A of rank r into sum of r matrices of rank 1. D. Leykekhman Linear Least Squares 5

8 Applications of SVD Decomposition. The 2-norm and the Frobenius norm of A can be easily computed from the SVD decomposition Ax 2 A 2 = sup = σ 1 x 0 x 2 m n A F = a 2 ij = σ σ2 p, j=1 p = min {m, n}. From the SVD decomposition of A it also follows that A T A = V Σ T ΣV T and AA T = UΣΣ T U T. Thus, σi 2, i = 1,..., p are the eigenvalues of symmetric matrices A T A and AA T and v i and u i are the corresponding eigenvectors. D. Leykekhman Linear Least Squares 6

9 Applications of SVD Decomposition. Theorem Let the SVD of A R m n be given by with r = rank(a). If k < r then A = U r Σ r V T r = A k = u i vi T k u i vi T, min A D 2 = A A k 2 = σ k+1, rank(d)=k and min A D p F = A A k F = σi 2, rank(d)=k k+1 D. Leykekhman Linear Least Squares 7 p = min {m, n}.

10 Solving LLS with SVD Decomposition. Consider the LLS min x Ax b 2 2 Let A = UΣV T be the SVD of A R m n. Using the orthogonality of U and V we have Ax b 2 2 = U T (AV V T x b) 2 2 = Σ V }{{} T x U T b) 2 2 = ( z i u T i b) 2 + =z m (u T i b) 2. i=r+1 D. Leykekhman Linear Least Squares 8

11 Solving LLS with SVD Decomposition. Thus, min x Ax b 2 2 = ( z i u T i b) 2 + m (u T i b) 2. i=r+1 D. Leykekhman Linear Least Squares 9

12 Solving LLS with SVD Decomposition. Thus, min x Ax b 2 2 = The solution is given ( z i u T i b) 2 + m (u T i b) 2. i=r+1 z i = ut i b, i = 1,..., r, z i = arbitrary, i = r + 1,..., n. D. Leykekhman Linear Least Squares 9

13 Solving LLS with SVD Decomposition. Thus, min x Ax b 2 2 = The solution is given ( z i u T i b) 2 + m (u T i b) 2. i=r+1 z i = ut i b, i = 1,..., r, z i = arbitrary, i = r + 1,..., n. As a result min x Ax b 2 2 = m (u T i b) 2. i=r+1 D. Leykekhman Linear Least Squares 9

14 Solving LLS with SVD Decomposition. Recall that z = V T x. Since V is orthogonal, we find that x 2 = V V T x 2 = V T x 2 = z 2. All solutions of the linear least squares problem are given by z = V T x with z i = ut i b, i = 1,..., r, z i = arbitrary, i = r + 1,..., n. D. Leykekhman Linear Least Squares 10

15 Solving LLS with SVD Decomposition. Minimum norm solution The minimum norm solution of the linear least squares problem is given by x = V z, where z R n is the vector with entries z i = ut i b, i = 1,..., r, z i The minimum norm solution is = 0, i = r + 1,..., n. x = u T i b v i D. Leykekhman Linear Least Squares 11

16 Solving LLS with SVD Decomposition. MATLAB code. % compute the SVD: [U,S,V] = svd(a); s = diag(s); % determine the effective rank r of A using singular values r = 1; while( r < size(a,2) & s(r+1) >= max(size(a))*eps*s(1) ) r = r+1; end d = U *b; x = V* ( [d(1:r)./s(1:r); zeros(n-r,1) ] ); D. Leykekhman Linear Least Squares 12

17 Conditioning of a Linear Least Squares Problem. Suppose that the data b are b = b ex + δb, where δb represents the measurement error. The minimum norm solution of min Ax (b ex + δb) 2 2 is x = u T i b v i = ( u T i b + ut i δb ) v i. D. Leykekhman Linear Least Squares 13

18 Conditioning of a Linear Least Squares Problem. Suppose that the data b are b = b ex + δb, where δb represents the measurement error. The minimum norm solution of min Ax (b ex + δb) 2 2 is x = u T i b v i = ( u T i b + ut i δb ) v i. If a singular value is small, then ut i (δb) could be large, even if u T i (δb) is small. This shows that errors δb in the data can be magnified by small singular values. D. Leykekhman Linear Least Squares 13

19 Conditioning of a Linear Least Squares Problem. % Compute A t = 10.^(0:-1:-10) ; A = [ ones(size(t)) t t.^2 t.^3 t.^4 t.^5]; % compute SVD of A [U,S,V] = svd(a); sigma = diag(s); % compute exact data xex = ones(6,1); bex = A*xex; for i = 1:10 % data perturbation deltab = 10^(-i)*(0.5-rand(size(bex))).*bex; b = bex+deltab; % solution of perturbed linear least squares problem w = U *b; x = V * (w(1:6)./ sigma); errx(i+1) = norm(x - xex); errb(i+1) = norm(deltab); end loglog(errb,errx, * ); ylabel( x^{ex} - x _2 ); xlabel( \delta b _2 ) D. Leykekhman Linear Least Squares 14

20 Conditioning of a Linear Least Squares Problem. The singular values of A in the above Matlab example are: σ σ σ σ σ σ The error x ex x 2 for different values of δb 2 (loglog-scale): We see that small perturbations δb in the measurements can lead to large errors in the solution x of the linear least squares problem if the singular values of A are small. D. Leykekhman Linear Least Squares 15

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