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Unformatted text preview: 18.06 Problem Set 1 Solution Due Wednesday, 11 February 2009 at 4 pm in 2106. Total: 145 points Problem 1: If k ~v k = 7 and k ~w k = 3, what are the smallest and largest possible values of k ~v + ~w k and ~v ~w ? Solution (10 points) k ~v + ~w k k ~v k + k ~w k = 10. k ~v + ~w k k ~v k  k ~w k = 4. (5pts)  ~v ~w  k ~v k k ~w k = 21. So 21 ~v ~w 21. (5pts) The maximum is achieved when the vectors ~v and ~w are parallel and pointing to the same direction, for example, ~v = (7 , , ,... ) and ~w = (3 , , ,... ); the minimum is achieved when they are parallel but pointing to opposite directions, for instance, ~v = (7 , , ,... ) and ~w = ( 3 , , ,... ). REMARK: We should try not to restrict ourselves to the 3dimensional case. The statement of this problem works for vectors in arbitrary dimensional space. Problem 2: Let A and B be 4 4 matrices, and divide each of them into 2 2 chunks via A = A 1 A 2 A 3 A 4 and B = B 1 B 2 B 3 B 4 , where A 1 is the upperleft 2 2 corner, A 2 is the upperright 2 2 corner, and so on. Let C = AB be the 4 4 product of A and B , and similarly divide C into 2 2 chunks as C = C 1 C 2 C 3 C 4 . (a) Give formulas for these 2 2 chunks C 1 ... 4 in terms of matrix products and sums of the chunks A 1 ... 4 and B 1 ... 4 (your final formulas should not reference the individual numbers within those chunks). (b) Justify your formulas by an example (come up with 4 4 matrices A and B with nonzero entries, multiply them to get C , and compare to your formulas in terms of 2 2 chunksit is acceptable to check just one of the output 2 2 chunks, say C 2 ). Solution (15 points = 10+5) (a) Similar to usual matrix multiplication, matrix multiplication by blocks for mally has the same form. C 1 = A 1 B 1 + A 2 B 3 , C 2 = A 1 B 2 + A 2 B 4 , C 3 = A 3 B 1 + A 4 B 3 , C 4 = A 3 B 2 + A 4 B 4 . 1 In other words, A 1 A 2 A 3 A 4 B 1 B 2 B 3 B 4 = A 1 B 1 + A 2 B 3 A 1 B 2 + A 2 B 4 A 3 B 1 + A 4 B 3 A 3 B 2 + A 4 B 4 NOTE: the order of A * and B * cannot be reversed. (b) For example, A = 1 0 0 1 0 1 1 1 1 1 0 1 0 1 1 0 , B = 0 1 0 1 1 1 0 1 1 0 1 1 1 0 1 0 . Then C = AB = 1 1 1 1 3 1 2 2 2 2 1 2 2 1 1 2 and C 1 = 1 0 0 1 0 1 1 1 + 0 1 1 1 1 0 1 0 = 0 1 1 1 + 1 0 2 0 = 1 1 3 1 , C 2 = 1 0 0 1 0 1 0 1 + 0 1 1 1 1 1 1 0 = 0 1 0 1 + 1 0 2 1 = 1 1 2 2 , C 3 = 1 1 0 1 0 1 1 1 + 0 1 1 0 1 0 1 0 = 1 2 1 1 + 1 0 1 0 = 2 2 2 1 , C 4 = 1 1 0 1 0 1 0 1 + 0 1 1 0 1 1 1 0 = 0 2 0 1 + 1 0 1 1 = 1 2 1 2 . Problem 3: Invent a 3 3 magic matrix M 3 with entries 0 , 1 ,..., 8, such that all rows and columns and diagonals add to 12 (e.g. the first row could be 7,2,3)....
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 Spring '08
 Strang

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