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Unformatted text preview: R ( T ) and a basis for R ( U ), the set { b + , + g : b , g } spans R ( T ) + R ( U ), and so the dimension of R ( T ) + R ( U ) is bounded by rank( T ) plus rank( U ). Then since R ( T + U ) is a subspace by part (a) the result follows. (c) . ..... 3.3 2. (b) 1 3 2 3 1 3. (a) Other vectors besides (5 , 0) would work (e.g., (2 , 1)). (b) 2 3 1 3 + a 1 3 2 3 1 : a R 4. (a) A1 = 5 3 21 , solution 11 5 6. (11 / 2 ,9 / 2 , 0) could be replaced, e.g. by (0 , 1 ,11) 1...
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This note was uploaded on 11/30/2009 for the course MATH 115A taught by Professor Liu during the Winter '07 term at UCLA.
 Winter '07
 Liu
 Linear Algebra, Algebra, Vector Space

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