08bAss3sol

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

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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 + { 0 } = 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 = { 0 } and U W = { 0 } , but since V + W = R 2 , we have U ( V + W ) = U . 1

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Let B = { u 1 , · · · , u r } be a basis for U V . (i.e. dim( U V ) = r .) Then { u 1 , · · · , u r } is a set of linearly independent vectors in U . By Theorem 3.4.4 (ii), let dim U = n , we can extend u 1 , · · · , u r by adding suitable vectors to form a basis for U . Let { u 1 , · · · , u r , b 1 , · · · , b n - r } be such a basis for U . (In case
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