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# Lecture10 - .1 Projection Matrix(Conti The formula for the...

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1 Lecture 10 10.1 Projection Matrix (Conti.) The formula for the projection matrix Example: Find the projection b = [ 1, 1, -2] on W = sp([1, -1, 2], [1, 0, -2]).

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2 Properties of P: 1. P 2 = P 2. P T = P. Let k q q q ..., , , 2 1 be an orthonormal basis of W. Then P = AA T . Example: Find the projection b = [ 1, 1, -2] on W = sp([1, -1, 2], [1, 0, -2]). 10.2 The method of least squares “Fitting” a curve to points (data obtained in experimental work). A straight line: y = a + bx
3 A quadratic polynomial: y = a + bx +cx 2 A cubic polynomial: y = a + bx +cx 2 +dx If we have data (x 1, y 1 ), (x 2, y 2 ), … (x m , y m ), then according the form of the function, we have a linear system y 1 = r 0 + r 1 x 1 + r 2 x 1 2 + … + r k x 1 k y 2 = r 0 + r 1 x 2 + r 2 x 2 2 + … + r k x 2 k y m = r 0 + r 1 x m + r 2 x m 2 + … + r k x m k and it can be rewritten as b x A , where k m m k k x x x x x x A ... 1 ... 1 ... 1 2 2 1 1 m y y y b 2 1 k r r r x 1 0 k a a a ..., , , 1 0 b is a linear combination of k a a a ..., , , 1 0 if all the data fit the linear system. For a vector x to be a least squares solution of b x A , (minimize the error || A x b|| ), this vector x must satisfy W b x A where W = sp k a a a ..., , , 2 1 . i.e. A x = A(A T A) -1 A T b

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4 Example: According to Newton’s second law of motion, a body near the earth’s surface falls
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