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Unformatted text preview: R and u 6 S, v S and v 6 R. Consider the vector u + v . As we are assuming R S is a subspace, R S is closed under addition. Hence u + v R S and so u + v R or u + v S . However if u + v R , then v = ( u + v )u R , which is a contradiction; similarly if u + v S . Hence we have derived a contradiction on the asumption that R 6 S and S 6 R . Consequently at least one of these must be false. In other words R S or S R . 19. Let X 1 ,...,X r be a basis for S . (i) irst let Y 1 = a 11 X 1 + + a 1 r X r . . . (2) Y r = a r 1 X 1 + + a rr X r , 40...
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 Fall '10
 Dreibelbis

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