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Unformatted text preview: as T ( a 1 v 1 + + a n v n ) = 0 . Hence a 1 v 1 + + a n v n null T . But T is injective, so null T = { } . Thus a 1 v 1 + + a n v n = 0 . Since v 1 ,...,v n are linearly independent, a i = 0 for each i as was to be shown. Date : July 2, 2009....
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 Summer '08
 GUREVITCH
 Math, Vector Space

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