Quiz 2 - 1 ,y 2 ,y 3 ) S . Thus, x 1 + x 2 + x 3 = 0 and y...

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Math 3A (44351) Quiz 2 Key Problem 1. Let S be the set of all ordered pairs of real numbers ( x 1 ,x 2 ). We define scalar multiplication and addition on S by: α * ( x 1 ,x 2 ) = ( αx 1 , - αx 2 ) ( x 1 ,x 2 ) ( y 1 ,y 2 ) = ( x 2 1 + y 2 1 ,x 2 2 + y 2 2 ) Show that S , with these operations, is not a vector space. Solution: Axiom A8 is not satisfied: there is no scalar ”1” s.t. 1 * ( x 1 ,x 2 ) = ( x 1 ,x 2 ) ( x 1 ,x 2 ) S , since, for any scalar α , α * ( x 1 ,x 2 ) = ( αx 1 , - αx 2 ) = ( x 1 ,x 2 ) iff αx 1 = x 1 and - αx 2 = x 2 for all x 1 ,x 2 R . Clearly, this means that α = 1 and α = - 1, which cannot both be true. Thus no such ”1” scalar exists, so Axiom A8 does not hold, and thus S , under these operations, is not a vector space. ± Problem 2. Prove that the set S = { ( x 1 ,x 2 ,x 3 ) T | x 1 + x 2 + x 3 = 0 } is a subspace of R 3 . Solution: We must show that it is nonempty, closed under addition, and closed under scalar multiplication. nonempty: Clearly (0 , 0 , 0) S and so S is nonempty. X closed under addition: Let x = ( x 1 ,x 2 ,x 3 ) S and y = ( y
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Unformatted text preview: 1 ,y 2 ,y 3 ) S . Thus, x 1 + x 2 + x 3 = 0 and y 1 + y 2 + y 3 = 0. We want to show that x + y = ( x 1 + y 1 ,x 2 + y 2 ,x 3 + y 3 ) S . Well this is true only if ( x 1 + y 1 ) + ( x 2 + y 2 ) + ( x 3 + y 3 ) = 0, which clearly true because ( x 1 + y 1 )+( x 2 + y 2 )+( x 3 + y 3 ) = ( x 1 + x 2 + x 3 )+( y 1 + y 2 + y 3 ) = 0+0 = 0. Thus S is closed under addition. X closed under scalar multiplication: Let x = ( x 1 ,x 2 ,x 3 ) S and let be scalar. Thus, x 1 + x 2 + x 3 = 0 and R . We want to show that x = ( x 1 ,x 2 ,x 3 ) S . This is true only if x 1 + x 2 + x 3 = 0, which is true because x 1 + x 2 + x 3 = ( x 1 + x 2 + x 3 ) = 0 = 0 . X Thus, S is a subspace of R 3 . 1...
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