08bAss3sol - 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. In lecture, we proved that W = Span( W ) if and only if W is a subspace. Let me repeat the argument below. [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.] Now we prove the result in Qn 1. For any nonempty subset S ⊂ V , Span( S ) is a subspace. Remains 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 . 1. 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 ). 2. No. Consider U = Span((1 1) T ), V = Span( e 1 ), W = Span( e 2 ), all of which are sub- spaces 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. 3. No. Use the counterexample in (b). Now, U ∩ V = { } and U ∩ W = { } , but since V + W = R 2 , we have U ∩ ( V + W ) = U ....
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This note was uploaded on 12/06/2010 for the course MATH MATH101 taught by Professor Chan during the Spring '09 term at HKUST.

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08bAss3sol - AS3sol/MATH1111/YKL/08-09 THE UNIVERSITY OF...

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