HW01_solution - 1. a + a = a (0 + 0 ) by (VS7) = a by (VS3)...

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Unformatted text preview: 1. a + a = a (0 + 0 ) by (VS7) = a by (VS3) = a + 0 by (VS3) by theorem 1.1 ⇒ a = 0 2. (1) It is not a vector space, since the conditions (VS2) (VS3) (VS4) (VS8) do not hold. (VS2) ∃ ( a 1 ,a 2 ) , ( b 1 ,b 2 ) , ( c 1 ,c 2 ) ∈ V (( a 1 ,a 2 )+( b 1 ,b 2 ))+( c 1 ,c 2 ) = ( a 1 + b 1 ,- a 2- b 2 )+( c 1 ,c 2 ) = ( a 1 + b 1 + c 1 ,a 2 + b 2- c 2 ) , ( a 1 ,a 2 )+(( b 1 ,b 2 )+( c 1 ,c 2 )) = ( a 1 ,a 2 )+( b 1 + c 1 ,- b 2- c 2 ) = ( a 1 + b 1 + c 1 ,- a 2 + b 2 + c 2 ) ⇒ (( a 1 ,a 2 ) + ( b 1 ,b 2 )) + ( c 1 ,c 2 ) 6 = ( a 1 ,a 2 ) + (( b 1 ,b 2 ) + ( c 1 ,c 2 )) (VS3) 0 does not exist. supposed exist, let 0 = ( x 1 ,x 2 ) ∈ V ( a 1 ,a 2 ) + ( x 1 ,x 2 ) = ( a 1 + x 1 ,- a 2- x 2 ) = ( a 1 ,a 2 ) ⇒ ( x 1 ,x 2 ) = (0 ,- 2 a 2 ) similarly choose ( b 1 ,b 2 ) ∈ V and b 2 6 = a 2 we get ( x 1 ,x 2 ) = (0 ,- 2 b 2 ) 6 = (0 ,- 2 a 2 ) = ( x 1 ,x 2 ) contradiction ⇒ does not exist. (VS4) because 0 does not exist. (VS8) ∃ ( a 1 ,a 2 ) ∈ V and c 1 ,c 2 ∈ R ( c 1 + c 2 )( a 1 ,a 2 ) = (( c 1 + c 2 ) a 1 ,a 2 ) = ( c 1 a 1 + c 2 a 1 ,a 2 ) c 1 ( a 1 ,a 2 ) + c 2 ( a 1 ,a 2 ) = ( c 1 a 1 ,a 2 ) + ( c 2 a 1 ,a 2 ) = ( c 1 a 1 + c 2 a 1 ,- 2 a 2 ) ⇒ ( c 1 + c 2 )( a 1 ,a 2 ) 6 = c 1 ( a 1 ,a 2 ) + c 2 ( a 1 ,a 2 ) (2) It is not a vector space, since the conditions (VS5) (VS6) (VS7) (VS8) do not hold. (VS5) ∃ ( a 1 ,a 2 ) ∈ V 1( a 1 ,a 2 ) = ( a 1 + 1 ,a 2 + 1) 6 = ( a 1 ,a 2 ) (VS6) ∃ ( a 1 ,a 2 ) ∈ V and c 1 ,c 2 ∈ < ( c 1 c 2 )( a 1 ,a 2 ) = ( c 1 c 2 a 1 + 1 ,c 1 c 2 a 2 + 1) c 1 ( c 2 ( a 1 ,a 2 )) = c 1 ( c 2 a 1 + 1 ,c 2 a 2 + 1) = ( c 1 c 2 a 1 + c 1 + 1 ,c 1 c 2 a 2 + c 1 + 1) ⇒ ( c 1 c 2 )( a 1 ,a 2 ) 6 = c 1 ( c 2 ( a 1 ,a 2 )) (VS7) ∃ ( a 1 ,a 2 ) , ( b 1 ,b 2 ) ∈...
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This note was uploaded on 11/09/2011 for the course EE EE taught by Professor Ee during the Spring '11 term at National Chiao Tung University.

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HW01_solution - 1. a + a = a (0 + 0 ) by (VS7) = a by (VS3)...

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