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Unformatted text preview: MATH 102 SOLUTIONS TO HW #2 Section 2.1, problem 2. The answers are as follows: a) This is as subspace. It consists of all vectors in R 3 of the form (0 , b 2 , b 3 ). Notice that setting b 2 = b 3 = 0 gives the zero vector. Also, this set is closed under addition because: (0 , b 2 , b 3 ) + (0 , c 2 , c 3 ) = (0 , b 2 + c 2 , b 3 + b 3 ). Likewise, we have that c (0 , b 2 , b 3 ) = (0 , cb 2 , cb 3 ), which is of the correct general form. b) This set is not a subspace. It consists of all vectors in R 3 of the form (1 , b 2 , b 3 ). Notice that the vector ~ 0 = (0 , , 0) cannot be written in this form. It is not difficult to see that the other vectorspace axioms (rules) are also violated for this set. c) This set is not a subspace. It consists of all vectors which are either of the form ( b 1 , , b 3 ) or ( b 1 , b 2 , 0). The problem is that if one adds two such vectors, say one of each form, then the result will not necessarily be of the same form (i.e. like one or the other). For example, let ~x = (1 , , 1) and ~ y = (1 , 1 , 0). Then ~ z = ~x + ~ y = (2 , 1 , 1) which is not of the form: either ( b 1 , , b 3 ) or ( b 1 , b 2 , 0). d) The set of all linear combinations of two vectors is always a subspace. For example, in the case of this problem two such combinations may be added together as follows: ( a 1 (1 , 1 , 0)+ a 2 (2 , , 1) ) + ( b 1 (1 , 1 , 0)+ b 2 (2 , , 1) ) = ( a 1 + b 1 )(1 , 1 , 0)+( a 2 + b 2 )(2 , , 1) . Therefore, we see that combinations are closed under addition. It is also clear that the other axioms of a linear subspace are satisfied. e) Solutions to a homogeneous linear equation are always a linear subspace. In this case, let ( b 1 , b 2 , b 3 ) and ( c 1 , c 2 , c 3 ) be two solutions to the equation in the text. Then a simple computation shows that their sum also satisfies the same equation: ( b 3 + c 3 ) ( b 2 + c 2 ) + 3( b 1 + c 1 ) = ( b 3 b 2 + 3 b 1 ) + ( c 3 c 2 + 3 c 1 ) = 0 + 0 = 0 ....
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 Fall '08
 SZYPOWSK
 Linear Algebra, Algebra, Vectors

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