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Unformatted text preview: MATH 223, Linear Algebra Winter 2010 Solutions to Assignment 2 C stands for the set of complex numbers, and R for the set of real numbers. 1. Here are some subsets of the complex vector space C 3 . In each case, decide whether the given set S is a subspace of C 3 ; justify your answers. (a) S = { a b 2 ib (1 + i ) a 2 b : a,b any complex numbers } . Solution: This is a subspace. First ~ ∈ S since we can take a = b = 0. Next, suppose that ~v 1 and ~v 2 are in S . Then ~v 1 = a 1 b 1 2 ib 1 (1 + i ) a 1 2 b 2 for some complex numbers a 1 and b 1 , and ~v 2 = a 2 b 2 2 ib 2 (1 + i ) a 2 2 b 2 for some complex a 2 , b 2 . So ~v 1 + ~v 2 = a 1 b 1 + a 2 b 2 2 ib 1 + 2 ib 2 (1 + i ) a 1 2 b 1 + (1 + i ) a 2 2 b 2 = ( a 1 + a 2 ) ( b 1 + b 2 ) 2 i ( b 1 + b 2 ) (1 + i )( a 1 + a 2 ) 2( b 1 + b 2 ) , which is of the right form to be an element of S . Hence S is closed under addition. Finally, suppose ~v ∈ S and α is a (complex) scalar. Then ~v = a b 2 ib (1 + i ) a 2 b for some complex numbers a and b , so α~v = ( αa ) ( αb ) 2 i ( αb ) (1 + i )( αa ) 2( αb ) , again an element of S . Thus S is closed under scalar multiplication. ( S is in fact Span { 1 1 + i ,  1 2 i 2 } . ) (b) S = { a b 2 ib + 2 (1 + i ) a 2 b : a,b any complex numbers } . Solution: This is not a subspace. In fact, ~ / ∈ S . If it were (note the use of the subjunctive) we would need a b = 2 ib +2 = (1+ i ) a 2 b = 0 1 for some complex a and b . The first two equations would tell us a = b = i , but then (1+ i ) a 2 b 6 = 0. So it’s impossible for S to have the zero vector. (In fact, it is not closed under either addition or scalar multiplication, either. But you don’t need to show that now.) (c) S = { a b 2 ib (1 + i ) a 2 b : a,b any real numbers } ....
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 Spring '07
 Loveys
 Linear Algebra, Algebra, Real Numbers, Vector Space, Complex Numbers, Sets, Complex number, scalar multiplication

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