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# ì„ ëŒ€1-2,3 - Linear...

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Linear Algebra I Solutions of Exercise 1.2 #1,8,13,17,18,19,21 1. (a) True. (b) False. (c) False. If x is the zero vector then this is false. (d) False. If a = 0 then this is false. (e) True. (f) False. (g) False. (h) False. If f ( x ) = x 2 + 1 and g ( x ) = - x 2 + 1 then deg f = deg g =2. But deg( f + g )= 0. (j) True. (k) True. 8. ( a + b )( x + y ) = ( a + b ) x + ( a + b ) y (by VS 7) = ax + bx + ay + by (by VS 8) = ax + ay + bx + by (by VS 1) 13. Let ( x, y ) be the zero vector. Then by VS 3, we must have: ( a, b ) + ( x, y ) = ( a, b ) for all ( a, b ) in V . Since ( a, b ) + ( x, y ) = ( a + b, by ) must be equal to ( a, b ), we have ( x, y ) = (0 , 1). On the other hand, by VS 4, for any vector ( a, b ) there must be a vector ( a , b ) such that ( a, b ) + ( a , b ) = (0 , 1). Since ( a, b ) + ( a , b ) = ( a + a , bb ) we have a = - a and b = 1 b . Hence if we take b = 0 then for any a , ( a, 0) has no such ( a , b ). Therefore, V is not a vector space. 17. By Theorem 1.2 (a), 0( a, b ) = ( a, 0) must be the zero vector. But then ( x, y )+( a, 0) = ( x + a, y ) = ( x, y ). Hence ( a, 0) cannot be the zero vector. This shows V is not a vector space under the operations. 18. ( a 1 , a 2 ) + ( b 1 , b 2 ) = ( a 1 + 2 b 1 , a 2 + 3 b 2 ) and ( b 1 , b 2 ) + ( a 1 , a 2 ) = ( b 1 + 2 a 1 , b 2 + 3 a 2 ). Hence ( a 1 , b 1 )+( a 2 , b 2 ) = ( a 2 , b 2 )+( a 1 , b 1 ) which says that our operation violates VS1. Accordingly, V is not a vector space. 19. Let c, d be nonzero scalars such that c + d = 0. Write x = ( a 1 , a 2 ). Then ( c + d ) x = (( c + d ) a 1 , a 2 c + d ) and cx + dx = (( c + d ) a 1 , a 2 ( 1 c + 1 d )). Therefore ( c + d ) x = cx + dx . Therefore V is not a vector space.

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