Using this weird addition again we have 0 0 1 0 2 0

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Using this weird addition again, we have (0 , 0) + (1 , 0) = (2 , 0), and (1 , 0) + (0 , 0) = (1 , 0). Therefore, (VS 1) does not hold, and so V is not a vector space under these operations. § 1.2 # 21 Let V and W be vector spaces over a field F . Let Z = { ( v, w ) : v V and w W } . Define addition and scalar-multiplication component-wise. The most straightforward way to prove that Z is a vector space is directly verify that all of the axioms hold. But most of them hold automatically since they hold component- wise. For example, to verify (VS 1), we calculate ( v 1 , w 1 ) + ( v 2 , w 2 ) = ( v 1 + v 2 , w 1 + w 2 ) = ( v 2 + v 1 , w 2 + w 1 ) = ( v 2 , w 2 ) + ( v 1 , w 1 ) . Notice that in the first and third equalities we only used the definition of addition in Z . The middle equality follows since V and W are vector spaces satisfying (VS 1). All of the other axioms are exactly the same to verify, which we will now do without further comment.
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HOMEWORK ASSIGNMENT 1 SOLUTIONS 3 (VS 2) (( v 1 , w 1 ) + ( v 2 , w 2 )) + ( v 3 , w 3 ) = ( v 1 + v 2 , w 1 + w 2 ) + ( v 3 , w 3 ) = (( v 1 + v 2 ) + v 3 , ( w 1 + w 2 ) + w 3 ) = ( v 1 + ( v 2 + v 3 ) , w 1 + ( w 2 + w 3 )) = ( v 1 , w 1 ) + ( v 2 + v 3 , w 2 + w 3 ) = ( v 1 , w 1 ) + (( v 2 , w 2 ) + ( v 3 , w 3 )) . (VS 3), set ~ 0 = (0 , 0). ( v, w ) + (0 , 0) = ( v, w ) . (VS 4) ( v, w ) + ( - v, - w ) = ( v + ( - v ) , w + ( - w )) = (0 , 0) . (VS 5) 1( v, w ) = (1 v, 1 w ) = ( v, w ) . (VS 6) ( ab )( v, w ) = (( ab ) v, ( ab ) w ) = ( a ( bv ) , a ( bw )) = a ( bv, bw ) = a ( b ( v, w )) . (VS 7) a (( v 1 , w 1 ) + ( v 2 , w 2 )) = a ( v 1 + v 2 , w 1 + w 2 ) = ( a ( v 1 + v 2 ) , a ( w 1 + w 2 )) = ( av 1 + av 2 , aw 1 + aw 2 ) = ( av 1 , aw 1 ) + ( av 2 , aw 2 ) = a ( v 1 , w 1 ) + a ( v 2 , w 2 ) . (VS 8) ( a + b )( v, w ) = (( a + b ) v, ( a + b ) w ) = ( av + bv, aw + bw ) = ( av, aw ) + ( bv, bw ) = a ( v, w ) + b ( v, w ) . § 1.3 # 9 Let W 1 , W 3 , and W 4 be the vector spaces described in exercise 8. We first describe W 1 W 3 . Let P = ( a 1 , a 2 , a 3 ) be some point inside of this intersection. Then since P W 1 we must have a 1 = 3 a 2 and a 3 = - a 2 . Therefore we can write P = (3 a 2 , a 2 , - a 2 ). Also, because P W 3 we must have 2 a 1 - 7 a 2 + a 3 = 0. Replacing a 1 and a 3 as above, this means we must have 0 = 2 a 1 - 7 a 2 + a 3 = 2(3 a 2 ) - 7 a 2 + ( - a 2 ) = - 2 a 2 . Therefore a 2 = 0 and hence P = (0 , 0 , 0). Clearly (0 , 0 , 0) is actually in W 1 W 3 , and this is the only possible point in the intersection, so W 1 W 3 = { (0 , 0 , 0) } . This is the zero subspace of R 3 .
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