Least squares Solution of Homogeneous Equations
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1 Least squares Solution of Homogeneous Equations supportive text for teaching purposes Revision: 1.2, dated: December 15, 2005 Tomáš Svoboda Czech Technical University, Faculty of Electrical Engineering Center for Machine Perception, Prague, Czech Republic
2 Introduction We want to find a n 1 vector h satisfying 2/5 Ah = 0, where A is m n matrix, and 0 is n 1 zero vector. Assume m n, and rank(a) = n. We are obviously not interested in the trivial solution h = 0 hence, we add the constraint h = 1. Constrained least squares minimization: Find h that minimizes Ah subject to h = 1.
3 Derivation I Lagrange multipliers h = argminh Ah subject to h = 1. We rewrite the constraint as 1 h h = 0 3/5
4 Derivation I Lagrange multipliers h = argminh Ah subject to h = 1. We rewrite the constraint as 1 h h = 0 3/5 To find an extreme (the sought h) we must solve h ( h A Ah + λ(1 h h) ) = 0.
5 Derivation I Lagrange multipliers h = argminh Ah subject to h = 1. We rewrite the constraint as 1 h h = 0 3/5 To find an extreme (the sought h) we must solve h ( h A Ah + λ(1 h h) ) = 0. We derive: 2A Ah 2λh = 0.
6 Derivation I Lagrange multipliers h = argminh Ah subject to h = 1. We rewrite the constraint as 1 h h = 0 3/5 To find an extreme (the sought h) we must solve h ( h A Ah + λ(1 h h) ) = 0. We derive: 2A Ah 2λh = 0. After some manipulation we end up with: (A A λe)h = 0 which is the characteristic equation. Hence, we know that h is an eigenvector of (A A) and λ is an eigenvalue.
7 Derivation I Lagrange multipliers h = argminh Ah subject to h = 1. We rewrite the constraint as 1 h h = 0 3/5 To find an extreme (the sought h) we must solve h ( h A Ah + λ(1 h h) ) = 0. We derive: 2A Ah 2λh = 0. After some manipulation we end up with: (A A λe)h = 0 which is the characteristic equation. Hence, we know that h is an eigenvector of (A A) and λ is an eigenvalue. The least-squares error is e = h A Ah = h λh.
8 Derivation I Lagrange multipliers h = argminh Ah subject to h = 1. We rewrite the constraint as 1 h h = 0 3/5 To find an extreme (the sought h) we must solve h ( h A Ah + λ(1 h h) ) = 0. We derive: 2A Ah 2λh = 0. After some manipulation we end up with: (A A λe)h = 0 which is the characteristic equation. Hence, we know that h is an eigenvector of (A A) and λ is an eigenvalue. The least-squares error is e = h A Ah = h λh. The error will be minimal for λ = mini λi and the sought solution is then the eigenvector of the matrix (A A) corresponding to the smallest eigenvalue.
9 Derivation II SVD Let A = USV, where U is m n orthonormal, S is n n diagonal with descending order, and V is n n also orthonormal. 4/5
10 Derivation II SVD Let A = USV, where U is m n orthonormal, S is n n diagonal with descending order, and V is n n also orthonormal. From orthonormality of U, V follows that USV h = SV h and V h = h. 4/5
11 Derivation II SVD Let A = USV, where U is m n orthonormal, S is n n diagonal with descending order, and V is n n also orthonormal. From orthonormality of U, V follows that USV h = SV h and V h = h. Substitute y = V h. Now, we minimize Sy subject to y = 1. 4/5
12 Derivation II SVD Let A = USV, where U is m n orthonormal, S is n n diagonal with descending order, and V is n n also orthonormal. From orthonormality of U, V follows that USV h = SV h and V h = h. Substitute y = V h. Now, we minimize Sy subject to y = 1. Remember that S is diagonal and the elements are sorted descendently. Than, it is clear that y = [0, 0,..., 1]. 4/5
13 Derivation II SVD Let A = USV, where U is m n orthonormal, S is n n diagonal with descending order, and V is n n also orthonormal. From orthonormality of U, V follows that USV h = SV h and V h = h. Substitute y = V h. Now, we minimize Sy subject to y = 1. Remember that S is diagonal and the elements are sorted descendently. Than, it is clear that y = [0, 0,..., 1]. From substitution we know that h = Vy from which follows that sought h is the last column of the matrix V. 4/5
![Further reading Richard Hartley and Andrew Zisserman, Multiple View Geometry in computer vision, Cambridge University Press, 2003 (2nd edition), [Appendix A5] 5/5 Gene H. Golub and Charles F.](/docs-images/72/66386353/images/14-0.jpg "Van Loan, Matrix Computation, John Hopkins University Press, 1996 (3rd edition). Eric W. Weisstein. Lagrange Multiplier. From MathWorld A Wolfram Web Resource. http://mathworld.wolfram.")
14 Further reading Richard Hartley and Andrew Zisserman, Multiple View Geometry in computer vision, Cambridge University Press, 2003 (2nd edition), [Appendix A5] 5/5 Gene H. Golub and Charles F. Van Loan, Matrix Computation, John Hopkins University Press, 1996 (3rd edition). Eric W. Weisstein. Lagrange Multiplier. From MathWorld A Wolfram Web Resource. Eric W. Weisstein. Singular Value Decomposition. From MathWorld A Wolfram Web Resource.
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