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Unformatted text preview: 2.4. (2) R = T 1 Since T is invertible, ∃ T 1 : V → V ⇒ T 1 = T 1 1 V = T 1 ( TR ) = ( T 1 T ) R = IR = R ∴ R = T 1 11. T : P 3 ( R ) → M 2 × 2 ( R ) is linear by T ( f ) = f (1) f (2) f (3) f (4) ¶ We are going to show that T ( f ) = 0 ⇒ f = 0(the zero polynomial) In this case, f (1) = 0( i.e.f ( c ) = b ) ,f (2) = 0 ,f (3) = 0 ,f (4) = 0 ∴ ∀ f ( c i ) = 0 ,i = 0 , 1 , 2 , 3 ∴ f ( x ) = 3 ∑ i =0 b i f i ( x ) = 3 ∑ i =0 f ( c i ) f i ( x ) = 0 ∴ f is the zero polynomial ∴ T is onetoone 12. φ β ( v i ) = [ v i ] β = e i = (0 , ··· , 1 , ··· , 0) t [ φ β ] γ β = I n , when γ is the standard basis for F n ∴ φ β is an isomorphism or φ β : V → F n is onto ⇒ φ β is an isomorphism because dimV = dimF n 13. ∼ is an equivalence relation on the class of vector space over F (i) ∼ is reflexive 107 PNUMATH 2.4. ∀ V ∈ C ,V ∼ V ( ∵ ) I V : V → V s.t I V = v, ∀ v ∈ V is an isomorphism (ii) ∼ is symmetric If V ∼ W , then W ∼ V ( ∵ ) If T : V → W is isomorphic then ∃ T 1 : W → V is isomorphic ∴ W ∼ V (iii) ∼ is transitive If V ∼ W and W ∼ Z , then V ∼ Z ( ∵ ) Let T : V → W and U : W → Z are isomorphic, then UT is isomorphic 14. Let V = { a a + b c ¶  a,b,c ∈ F } T : V → F 3 s.t T a a + b c ¶ = ( a,b,c ) For the basis for V, { v 1 = 1 1 0 0 ¶ ,v 2 = 0 1 0 0 ¶ ,v 3 = 0 0 0 1 ¶ } ∃ ! T ( v i ) = w i is linear, w i ∈ F 3 i = 1 , 2 , 3 and since dimV = dimF 3 , V is isomorphic to F 3 ∴ T is an isomorphism from...
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 Spring '09
 RUDYAK
 Vector Space, basis, Trigraph, Equivalence relation, aki aij

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