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251 Lecture 18 LScondition

# 251 Lecture 18 LScondition - Least Squares General...

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Click to edit Master subtitle style 4/14/11 Least Squares General statement: For full rank A, m ≥ n, and m-vector b, find x such that || r || = || b – Ax || is minimized We have x = A+b where A+ = (A*A)-1A* is called the generalized inverse Nearest point to b is y = Ax given by y = Pb Where the orthogonal projector P = AA+

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Click to edit Master subtitle style 4/14/11 Least Squares Examples Nearest point: For points pi in R n, find p in R n such that f (pi) = ||pi - p ||2 = Σ(pi - p )2 is minimum.
Click to edit Master subtitle style 4/14/11 Least Squares Examples Nearest point: For points pi in R n, find p in R n such that f (pi) = ||pi - p ||2 = Σ(pi - p )2 is minimum, l Set δ f / δ pi = Σ 2(pi – p) δ pj/ δ pi = 0 à (pi – p) δ f / δ pi = 0, for all I à pi δ f / δ pi/ δ f / δ pi = p

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Click to edit Master subtitle style 4/14/11 Least Squares General statement: For full rank A, m ≥ n, and m-vector b, find x such that || r || = || b – Ax || is minimized b We have x = A+b and y = Pbr = b - Ax y = Ax = Pb angle θ
Click to edit Master subtitle style 4/14/11 Least Squares General statement: Re-draw picture We have x = A+b and y = Pb

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251 Lecture 18 LScondition - Least Squares General...

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