pset4-s10-soln

pset4-s10-soln - 18.06 Problem Set 4 Solution Due Thursday...

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Unformatted text preview: 18.06 Problem Set 4 Solution Due Thursday, 4 March 2010 at 4 pm in 2-106. Total: 100 points Section 3.5. Problem 2: (Recommended) Find the largest possible number of independent vectors among v 1 = 1- 1 v 2 = 1- 1 v 3 = 1- 1 v 4 = )0 1- 1 v 5 = )0 1- 1 v 6 = )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 A = 1- 2 3 The special solutions v 1 = 2 1 v 2 = - 3 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 basis for it. Finally, the vector v 3 = 1- 2 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: If a b c d e f g h i 0 1 0 0 0 1 0 0 0 = 0 1 0 0 0 1 0 0 0 a b c d e f g h i , then A = a b c a b 0 0 ai “The subspace of matrices that commute with the shift S has dimension .” pset4-s10-soln: page 2 Solution (4 points): Multiplying out both sides gives a b d e g h = d e f g h i 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....
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This note was uploaded on 02/22/2012 for the course EE 441 taught by Professor Neely during the Spring '08 term at USC.

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pset4-s10-soln - 18.06 Problem Set 4 Solution Due Thursday...

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