solutions2.1 - Math 307 Problems for section 2.1 October 4...

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Math 307: Problems for section 2.1 October 4, 2009 1. Are the vectors 1 2 1 2 1 , 1 0 2 1 1 , 1 1 3 2 0 , 0 0 2 0 1 , 0 4 9 7 3 linearly independent? You may use MAT- LAB/Octave to perform calculations, but explain your answer. Put the vectors in the columns of matrix and perform row reduction V1 = [1 2 1 2 1]’; V2 = [1 0 -2 1 1]’; V3 = [1 -1 3 -2 0]’; V4 = [0 0 -2 0 1]’; V5 = [0 4 -9 7 3]’; A=[V1 V2 V3 V4 V5]; rref(A) ans = 1.00000 0.00000 0.00000 0.00000 1.00000 0.00000 1.00000 0.00000 0.00000 1.00000 0.00000 0.00000 1.00000 0.00000 -2.00000 0.00000 0.00000 0.00000 1.00000 1.00000 0.00000 0.00000 0.00000 0.00000 0.00000 Since the matrix has rank 4 there are only 4 independent vectors in the list, and the vectors are linearly dependent. In fact a vector in the nullspace of A is 1 1 2 1 1 so V 1 V 2 + 2 V 3 V 4 + V 5 = 0 which we can check: -V1 - V2 + 2*V3 - V4 + V5 ans = 0 0 0 0 0 1
2. Which of the following sets are subspaces of the vector space V ? Why, or why not? (a) The set S = { ( b 1 , b 2 , b 3 ) : b 1 = 0; b 2 , b 3 R } . ( V = R 3 ) (b) The set S = { ( b 1 , b 2 , b 3 ) : b 1 b 2 = 0 , b 3 R } . This is union of the plane b 1 = 0 and the plane b 2 = 0 . ( V = R 3 ) (c) All infinite sequences ( x 1 , x 2 , . . . ) , with x i R and x j = 0 from some fixed point onwards. ( V = R ) (d) All non-increasing sequences ( x 1 , x 2 , . . . ) , with x i R and x j +1 x j for each j . ( V = R ) (e) The set of all polynomial functions, p ( x ) , where p ( x ) = 0 or p ( x ) has degree n for some fixed n 1 . ( V is the vector space of all polynomials.) (f) The set of odd continuous functions on the interval [ 1 , 1] , i.e., f C [ 1 , 1] such that f ( x ) = f ( x ) . ( V = C [ 1 , 1] ) (a) S is a subspace. The zero vector, (0 , 0 , 0), is in S . S is also closed under addition and scalar multiplication. Let x 1 = (0 , y 1 , z 1 ) and x 2 = (0 , y 2 , z 2 ) be two arbitrary vectors in S and let c R . Then x 1 + x 2 = (0 , y 1 + y 2 , z 1 + z 2 ) is also in S and c x 1 = (0 , cy 1 , cz 1 ) is also in S . (b) S is not a subspace. It is not closed under addition. Take for example the vectors (0 , 1 , 1) and (1 , 0 , 1). They are both in S but their sum, (1 , 1 , 2), is not in S . (c) This is a subspace. The zero vector clearly has x j = 0 from some point onwards. If we add two vectors which both have x j = 0 from some point onwards then the sum will also have x j = 0 from the same point onwards. Also, if we multiply a vector with x j = 0 from some point onwards by a scalar, the result will still have x j = 0 from the same point on. So the set is closed under addition and scalar multiplication. (d) This is not a subspace. It is not closed under scalar multiplication. Take for example the vector (2 , 1 , 0 , 0 , 0 , . . . ). If we multiply this by ( 1) we get ( 2 , 1 , 0 , 0 , 0 , . . . ), which is no longer non- increasing. (e) This is not a subspace. Take for example the two polynomials p 1 ( x ) = x n + 1 and p 2 ( x ) = x n .

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