1.6 Basis and Dimensions 2

# 1.6 Basis and Dimensions 2 - 1.6 6 f 3 x = 1 12 x 3 5 x 2 6...

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Unformatted text preview: 1.6. 6) , f 3 ( x ) = 1 12 ( x 3 + 5 x 2 + 6 x ) ∴ g ( x ) = 3 ∑ i =0 b i f i ( x ) =- 3 x 3- 6 x 2 + 4 x + 15 11. (i) We need to show that { u + v,au } is linearly independent k 1 ,k 2 are scalars, If k 1 ( u + v ) + k 2 ( au ) = ( k 1 + ak 2 ) u + ( k 1 v ) = 0 Since { u,v } is a basis for V , ∴ k 1 + ak 2 = 0 ,k 1 = 0 ( a is a nonzero scalar) ∴ k 1 = k 2 = 0 (ii) If k 1 ( au ) + k 2 ( bv ) = ( ak 1 ) u + ( bk 2 ) v = 0 Since a 6 = 0 ,b 6 = 0 and { u,v } is a basis for V . ∴ k 1 = k 2 = 0 ∴ { au,bv } is a basis for V . 12. k 1 ( u + v + W ) + k 2 ( v + w ) + k 3 ( w ) = 0 , k 1 ,k 2 ,k 3 ∈ F ⇒ k 1 u + ( k 1 + k 2 ) v + ( k 1 + k 2 + k 3 ) w = 0 ⇒ k 1 = k 2 = k 3 = 0 13. { a (1 , 1 , 1) | a ∈ R } is a solution set ∴ { (1 , 1 , 1) } is a basis for the given system 38 PNU-MATH 1.6. 14. { (0 , , 1 , , 0) , (0 , , , 1 , 0) , (0 , 1 , , , 0) , (0 , , , , 1) } : a basis for W 1 { (1 , , , ,- 1) , (0 , 1 , 1 , 1 , 0) } : a basis for W 2 ∴ dim( W 1 )=4, dim( W 2 )=2 (*) The dimension of the solution space AX = 0 is equal to n- rank A ( n is the number of rows of A ) (i) a 1- a 3- a 4 = 0 i.e. a 1 + 0 a 2- a 3- a 4 + 0 a 5 = 0—(*) A = (1 , ,- 1 ,- 1 , 0) 1 × 5 , X = a 1 a 2 a 3 a 4 a 5...
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## This note was uploaded on 02/14/2012 for the course MAS 4105 taught by Professor Rudyak during the Spring '09 term at University of Florida.

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1.6 Basis and Dimensions 2 - 1.6 6 f 3 x = 1 12 x 3 5 x 2 6...

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