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pset5-s09-soln

# pset5-s09-soln - 18.06 Problem Set 5 Solution Due Wednesday...

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18.06 Problem Set 5 Solution Due Wednesday, 18 March 2009 at 4 pm in 2-106. Total: 130 points. Problem 1: Write down three equations for the line b = C + Dt to go through b = 7 at t = - 1, b = 7 at t = 1, and b = 21 at t = 2. Find the least-squares solution ˆ ~x = ( C, D ) T . Sketch these three points and the line you found (or use a plotting program). Solution (10 points) The equations for the line b = C + Dt is 1 - 1 1 1 1 2 C D = 7 7 21 . Thus, the least-squares solution is given by solving 1 1 1 - 1 1 2 1 - 1 1 1 1 2 C D = 1 1 1 - 1 1 2 7 7 21 , 3 2 2 6 C D = 35 42 Hence, C = 9 , D = 4 and then ˆ ~x = (9 4) T . We plot the three points and the lines using MATLAB as follows, where the blue line is the line in the first problem and the red line is the one that passes origin in Problem 2. 2 1.5 1 0.5 0 0.5 1 1.5 2 2.5 3 15 10 5 0 5 10 15 20 25 t x 1

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>> plot(-1, 7, ’x’); plot(1, 7, ’x’); plot(2, 21, ’x’); >> hold on >> fplot(@(x) 4*x + 9, [-2, 3], ’b’); >> fplot(@(x) 7*x, [-2, 3], ’r’); >> xlabel(’t’); ylabel(’x’); >> grid Problem 2: For the same three points as in the previous problem, find the best-fit (least-squares) line through the origin . (What is the equation of a line through the origin? How many unknown parameters do you have?) Sketch this line on your plot from the previous problem. Solution (10 points) The equation for the line becomes b = Dt as the line goes through the origin. So we need to find the least-squares solution to the following linear system. - 1 1 2 ( D ) = 7 7 21 Similar to Problem 1, we need to solve ( - 1 1 2 ) - 1 1 2 ( D ) = ( - 1 1 2 ) 7 7 21 Hence, 6 D = 42, and we have D = 7. Problem 3: If we solve A T A ˆ ~x = A T ~ b , which of the four subspaces of A contains the error vector ~e = ~ b - A ˆ ~x ? Which contains the projection ~ p = A ˆ ~x ? Can ˆ ~x be chosen to lie completely inside any of the subspaces, and if so which one? Solution (15 points = 5+5+5) Since the error vector is orthogonal to the column space C ( A ), it lies in the left nullspace N ( A T ). The projection ~ p is on the column space C ( A ) (because it is the projection onto the column space). Since the row space C ( A T ) and the nullspace N ( A ) spans the whole space, and we can always modify the vector ˆ ~x by a vector in N ( A ) (which does not affect the projection A ˆ ~x ). Hence, we can choose ˆ ~x to be in the row space C ( A T ). 2
REMARK: We actually see this phenomenon in Problem 11(d) of Pset 4. Problem 4: In this problem, you will use 18.02-level calculus to understand why the solution to A T A~x = A T ~ b minimizes k A~x - ~ b k over all ~x , for any arbitrary m × n matrix A . Consider the function: f ( ~x ) = k A~x - b k 2 = ( A~x - ~ b ) T ( A~x - ~ b ) (1) = ~x T A T A~x - ~ b T A~x - ~x T A T ~ b + ~ b T ~ b (2) = X i,j B ij x i x j - 2 X i,j A ij b i x j + ~ b T ~ b, (3) where B = A T A . Compute the partial derivatives ∂f/∂x k (for any k = 1 , . . . , n ), and show that ∂f/∂x k = 0 (true at the minimum of f ) leads to the system of n equations A T A~x = A T ~ b .

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pset5-s09-soln - 18.06 Problem Set 5 Solution Due Wednesday...

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