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1.6 Basis and Dimensions 3

# 1.6 Basis and Dimensions 3 - 1.6 By(a v W1 and v W2 By the...

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1.6. ( ) By (a), v W 1 and v / W 2 By the exercise 20, dim( W 1 ) k and dim( W 2 ) k + 1 dim( W 1 ) dim( W 2 ) dim( W 1 ) < dim( W 2 ) 24. f ( x ) = k n x n + ··· + k 1 x + k 0 , k n 6 = 0 , k i R Let a 0 f ( x ) + a 1 f 0 ( x ) + ··· + a n f ( n ) ( x ) = 0, for some scalars a 0 , ··· ,a n R then ( a 0 k n ) x n + ( a 0 k n - 1 + a 1 k n ) x n - 1 + ··· + ( a 0 k 1 + 2! a 1 k 2 + 3! a 2 k 3 + ··· + n ! a n - 1 k n ) x + ( n m =0 m ! a m k m ) = 0 By equating the coeﬃcient of x k on both sides of this equation for k = 0 , 1 , 2 , ··· ,n , we obtain a 0 = a 1 = ··· = a n = 0 (since char R =0) It follows from (b) of corollary 2 (p.48) that { f ( x ) ,f 0 ( x ) , ··· ,f ( n ) ( x ) } is a basis for P n ( R ) . g ( x ) P n ( R ) , c 0 , ··· ,c n R s.t. g ( x ) = c 0 f ( x ) + c 1 f 0 ( x ) + ··· + c n f ( n ) ( x ) 25. Z = { ( v,w ) | v V and w W } = V × W dim( Z ) = dim( Z ) × dim( W ) = mn β = { v 1 , ··· ,v m } a basis for V γ = { w 1 , ··· ,w n } a basis for W then α = { ( v 1 , 0) , ··· , ( v m , 0) , ··· , (0 ,w 1 ) , ··· , (0 ,w n ) } a basis for V × W 44 PNU-MATH

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1.6. 26.
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1.6 Basis and Dimensions 3 - 1.6 By(a v W1 and v W2 By the...

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