081st_tut6sol

081st_tut6sol - MATH1111/2008-09/Tutorial VI Solution 1...

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Unformatted text preview: MATH1111/2008-09/Tutorial VI Solution 1 Tutorial VI Suggested Solution 1. Let V be a vector space and dim V = 2008, and let S,T be subspaces of V . Prove or disprove the following. (a) For every integer r with 0 ≤ r ≤ 2008, one can always find a subspace W r of V such that dim W r = r . (b) S = T if and only if dim S = dim T . (c) dim( S ∩ T ) ≤ min(dim S, dim T ) where min( a,b ) = the minimum of a and b . (d) dim( S ∪ T ) ≥ max(dim S, dim T ) where max( a,b ) = the maximum of a and b . (e) Let v ∈ V and U = Span( v ). Then dim( U + S ) = dim S + 1. Ans . Let { v 1 ,v 2 , ··· ,v 2008 } be a basis for V . (a) True. For r = 0, we take W = { } which is a subspace of V and dim W = 0. For 1 ≤ r ≤ 2008, we take W r to be the subspace Span( v 1 ,v 2 , ··· ,v r ). You can check easily that v 1 ,v 2 , ··· ,v r forms a basis for W r . So dim W r = r . (b) False. The ”only if” part is obviously correct (i.e. S = T are subspaces ⇒ dim S = dim T .) However the ”if” part is wrong. Consider V = R 2 and let S = Span( e 1 ) and T = Span( e 2 ). Both S and T are subspaces of V . And dim S = 1 = dim T but S 6 = T . (c) True. Note that S ∩ T is a subspace of V when S and T are subspaces of V . As S ∩ T ⊂ S , dim( S ∩ T ) ≤ dim S . (See lecture slides.) As S ∩ T ⊂ T , dim( S ∩ T ) ≤ dim T ....
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081st_tut6sol - MATH1111/2008-09/Tutorial VI Solution 1...

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