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# ì„ ëŒ€1-4,5 - 4 Solutions...

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4 Solutions of Exercise 1.4 #1, 2(b) (c) (e) (f), 3(a) (b) (c) 4(a), 6, 10, 12, 13, 14, 15 1. (a) True. (b) False. The span of is { 0 } . (c) True. We need to prove this. Let { W i } be the collection of all subspaces of V which contain S . Then i W i conatins S and being an intersection of subspaces it is again a subspace. Claim: i W i is the smallest subspace of V which contains S . In fact, if W is a subspace which contain S then W = W i for some i . Hence W ⊇ ∩ i W i , which says i W i is the smallest subspace of V which contain S . Therefore span( S ) = i W i . (d) False. The constant must be nonzero. (e) True. (f) False. 2. 3. (a)Yes. (b)No. (c) No. 4. (a) The first polynomial is a linear combination of the last two if and only if we can find a, b such that x 3 - 3 x +5 = a ( x 3 +2 x 2 - x +1)+ b ( x 3 +3 x 2 - 1). This is equivalent to the existence of the solution of the simultaneous linear equations a + b = 1 2 a + 3 b = 0 - a = - 3 a - b = 5 Now we see that the equation has solution a = 3 , b = - 2. 6. For any ( a, b, c ) F 3 we need to find x, y, z F for which ( a, b, c ) = x (1 , 1 , 0) + y (1 , 0 , 1) + z (0 , 1 , 1). Therefore we need to solve x + y = a x + z = b y + z = c The solution to this equation is x = b - c 2 , y = a - b + c 2 , z = - a + b 2 .

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