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08aAss3sol - 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. (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 + { 0 } = 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 = { 0 } and U W = { 0 } , 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. 1
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(b) Claim: X W = X . Proof. 1 X W X . This is obvious by the definition of intersection. 2 X X W . Let f X . Then f is of the form f = α x for some scalar α . As x = 2 u + 2 v , we see that f can be expressed as f = 2 α u + 2 α v .
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