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Linear Least Squares QR Factorization

Linear Least Squares QR Factorization. Systems of linear equations. Problem to solve: M x = b Given M x = b : Is there a solution? Is the solution unique?. Systems of linear equations. Find a set of weights x so that the weighted sum of the

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Linear Least Squares QR Factorization

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  1. Linear Least Squares QR Factorization

  2. Systems of linear equations • Problem to solve: M x = b • Given M x = b : • Is there a solution? • Is the solution unique?

  3. Systems of linear equations Find a set of weights x so that the weighted sum of the columns of the matrix M is equal to the right hand side b

  4. Systems of linear equations - Existence A solution exists when b is in the span of the columns of M A solution exists if: There exist weights, x1, …., xN, such that:

  5. Systems of linear equations - Uniqueness Suppose there exist weights, y1, …., yN, not all zero, such that: Then: Mx = b  Mx + My= b  M(x+y) = b A solution is unique only if the columns of M are linearlyindependent.

  6. QR factorization 1 • A matrix Q is said to be orthogonal if its columns are orthonormal, i.e. QT·Q=I. • Orthogonal transformations preserve the Euclidean norm since • Orthogonal matrices can transform vectors in various ways, such as rotation or reflections but they do not change the Euclidean length of the vector. Hence, they preserve the solution to a linear least squares problem.

  7. QR factorization 2 Any matrix A(m·n) can be represented as A = Q·R ,where Q(m·n) is orthonormal and R(n·n) is upper triangular:

  8. QR factorization 2 • Given A , let its QR decomposition be given as A=Q·R, where Q is an (m x n) orthonormal matrix and R is upper triangular. • QR factorization transform the linear least square problem into a triangular least squares. Q·R·x = b R·x = QT·b x=R-1·QT·b Matlab Code:

  9. Normal Equations Consider the system It can be a result of some physical measurements, which usually incorporate some errors. Since, we can not solve it exactly, we would like to minimize the error: r=b-Ax r2=rTr=(b-Ax)T(b-Ax)=bTb-2xTATb+xTATAx (r2)x=0 - zero derivative is a (necessary) minimum condition -2ATb+2ATAx=0; ATAx = ATb; – Normal Equations

  10. Normal Equations 2 ATAx = ATb – Normal Equations

  11. Least squares via A=QR decomposition A(m,n)=Q(m,n)R(n,n), Q is orthogonal, therefore QTQ=I. QRx=b R(n,n)x=QT(n,m)b(m,1) -well defined linear system x=R-1QTb Q is found by Gram=Schmidt orthogonalization of A. How to find R? QR=A QTQR=QTA, but Q is orthogonal, therefore QTQ=I: R=QTA R is upper triangular, since in orthogonalization procedure only a1,..ak (without ak+1,…) are used to produce qk

  12. Least squares via A=QR decomposition 2 Let us check the correctness: QRx=b Rx=QTb x=R-1QTb

  13. Last lecture reminderQR Factorization – By picture

  14. QR Factorization – Minimization ViewMinimization Algorithm For i = 1 to N “For each Target Column” For j = 1 to i-1 “For each Source Column left of target” end end Orthogonalize Search Direction Normalize

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