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MIT18_06S10_pset4_s10_soln

MIT18_06S10_pset4_s10_soln - 18.06 Problem Set 4 Solution...

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18.06 Problem Set 4 Solution Total: 100 points Section 3.5. Problem 2: (Recommended) Find the largest possible number of independent vectors among 1 1 1 )0 )0 )0 v 1 = 1 0 0 1 v 4 = 1 1 0 v 5 = 1 0 v 6 = 0 1 . 1 0 0 0 v 2 = v 3 = 0 1 1 Solution (4 points): Since v 4 = v 2 v 1 , v 5 = v 3 v 1 , and v 6 = v 3 v 2 , there are at most three independent vectors among these: furthermore, applying row reduction to the matrix [ v 1 v 2 v 3 ] gives three pivots, showing that v 1 , v 2 , and v 3 are independent. Section 3.5. Problem 20: Find a basis for the plane x 2 y + 3 z = 0 in R 3 . Then find a basis for the intersection of that plane with the xy plane. Then find a basis for all vectors perpendicular to the plane. Solution (4 points): This plane is the nullspace of the matrix 1 2 3 A = 0 0 0 0 0 0 The special solutions 2 3 0 0 1 give a basis for the nullspace, and thus for the plane. The intersection of this plane with the xy plane is a line: since the first vector lies in the xy plane, it must lie on the line and thus gives a 1 v 1 = v 2 = basis for it. Finally, the vector 1 2 v 3 = 3 is obviously perpendicular to both vectors: since the space of vectors perpendicular to a plane in R 3 is one-dimensional, it gives a basis. Section 3.5. Problem 37: If AS = SA for the shift matrix S , show that A must have this special form: a b c 0 1 0 0 1 0 a b c a b c d e f 0 0 1 = 0 0 1 d e f , then A = 0 a b If g h i 0 0 0 0 0 0 g h i 0 0 ai “The subspace of matrices that commute with the shift S has dimension .”
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pset4-s10-soln: page 2 Solution (4 points): Multiplying out both sides gives 0 a b d e f 0 d e = g h i 0 g h 0 0 0 Equating them gives d = g = h = 0 , e = i = a, f = b , i.e. the matrix with the form above. Since there are three free variables, the subspace of these matrices has dimension 3. Section 3.5. Problem 41: Write a 3 by 3 identity matrix as a combination of the other five permutation matrices. Then show that those five matrices are linearly indpendent. (Assume a combination gives c 1 P 1 + + c 5 P 5 = 0, and check entries to prove c i is zero.) The five permutation · · · matrices are a basis for the subspace of 3 by 3 matrices with row and column sums all equal. Solution (12 points): The other five permutation matrices are 1 1 1 1 1 P 21 = 1 , P 31 = 1 , P 32 = 1 , P 32 P 21 = 1 , P 21 P 32 = 1 1 1 1 1 1 Since P 21 + P 31 + P 32 is the all 1s matrix and P 32 P 21 + P 21 P 32 is the matrix with 0s on the diagonal and 1s elsewhere, I = P 21 + P 31 + P 32 P 32 P 21 P 21 P 32 . For the second part, the combination above gives c 3 c 1 + c 4 c 2 + c 5 c 1 + c 5 c 2 c 3 + c 4 = 0 c 2 + c 4 c 3 + c 5 c 1 Setting each element equal to 0 first
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