08aAss3sol

08aAss3sol - AS3sol/MATH1111/YKL/08-09 THE UNIVERSITY OF...

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Unformatted text preview: AS3sol/MATH1111/YKL/08-09 THE UNIVERSITY OF HONG KONG DEPARTMENT OF MATHEMATICS MATH1111: Linear Algebra Assignment 3 Suggested Solution 1. (a) No matter whether or not W is a subspace, W ⊂ Span( W ) since w ∈ W ⇒ 1 · w ∈ Span( W ). ”If part”: If W is a subspace, then any linear combination of elements of W belongs to W . i.e. Span( W ) ⊂ W . ∴ W = Span( W ). ”Only if” part: Clear by direct checking from definition. (b) As in part (a), we know S ⊂ Span( S ) and Span( S ) is a subspace. Want to show: S ⊂ W and W is a subspace ⇒ Span( S ) ⊂ W . This is also clear, since w 1 ,w 2 , ··· ,w r ∈ S ⇒ w 1 ,w 2 , ··· ,w r ∈ W . Thus, a 1 w 1 + a 2 w 2 + ··· + a r w r ∈ W because W is a subspace. ∴ Span( S ) ⊂ W . 2. (a) Since U + V is a subspace containing the sets U and V , by Qn 1 part (b), Span( U ∪ V ) ⊂ U + V . Let x ∈ U + V . Then x = u + v where u ∈ U and v ∈ V . So x is a linear combination of vectors in U ∪ V . i.e. U + V ⊂ Span( U ∩ V ). (b) No. Consider U = Span((1 1) T ), V = Span( e 1 ), W = Span( e 2 ), all of which are subspaces of R 2 . Then U + ( V ∩ W ) = U + { } = U but ( U + V ) ∩ ( U + W ) = R 2 ∩ R 2 = R 2 . Remark. U + ( V ∩ W ) ⊂ ( U + V ) ∩ ( U + W ) is always true. (c) No. Use the counterexample in (b). Now, U ∩ V = { } and U ∩ W = { } , but since V + W = R 2 , we have U ∩ ( V + W ) = U . 3. (a) dim X = 1 as { x } is a basis for X . Similarly, dim Y = 1. dim W = 2 as { u , v } is a basis for W . dim( X + W ) = 2 as X + W = { t + s : t ∈ X, s ∈ W } = { α x + β u + γ v : α,β,γ ∈ R } . However, x = 2 u + 2 v so X + W = Span( u , v ) where u , v are linearly independent. dim( Y + W ) = 3 as Y + W = Span( y , u , v ) and y , u , v are linearly independent....
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This note was uploaded on 09/06/2010 for the course MATH MATH1111 taught by Professor Forgot during the Fall '08 term at HKU.

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08aAss3sol - AS3sol/MATH1111/YKL/08-09 THE UNIVERSITY OF...

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