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Unformatted text preview: 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 − 1 ⎢ ⎢ ⎣ ⎥ ⎥ ⎦ ⎥ ⎥ ⎦ ⎢ ⎢ ⎣ ⎥ ⎥ ⎦ v 4 = ⎢ ⎢ ⎣ 1 − 1 ⎥ ⎥ ⎦ v 5 = ⎢ ⎢ ⎣ 1 ⎥ ⎥ ⎦ v 6 = ⎢ ⎢ ⎣ 1 ⎥ ⎥ ⎦ . − 1 v 2 = v 3 = − 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 = ⎣ ⎦ The special solutions ⎤ ⎡ ⎤ ⎡ 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 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 onedimensional, 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 1 1 a b c a b c ⎣ d e f ⎣ ⎦ 1 ⎦ = ⎣ 1 ⎣ ⎦ d e f ⎦ , then A = ⎣ a b ⎦ If g h i g h i ai “The subspace of matrices that commute with the shift S has dimension .” pset4s10soln: page 2 Solution (4 points): Multiplying out both sides gives ⎡ ⎤ ⎡ ⎤ a b d e f ⎣ d e ⎦ = ⎣ g h i ⎦ g h 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 08/22/2011 for the course MATH 1806 taught by Professor Strang during the Fall '10 term at MIT.
 Fall '10
 Strang
 Linear Algebra, Algebra, Vectors

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