advertisement

Math 2270 Quiz 7 1. In the space of all infinite sequences, consider the subspace W of arithmetic sequences, i.e., sequences of the form (a, a + k, a + 2k, a + 3k, . . .) for some constants a and k. Find a basis for W and verify that your basis satisfies all the requirements for a basis. If a sequence is in W , then it can be written as (a, a + k, a + 2k, a + 3k, . . .) = a(1, 1, 1, 1, . . .) + k(0, 1, 2, 3, . . .). Therefore, W = span((1, 1, 1, 1, . . .), (0, 1, 2, 3, . . .)). All we have left to do is show that these two sequences are linearly independent. Suppose c1 and c2 are constants such that c1 (1, 1, 1, 1, . . .) + c2 (0, 1, 2, 3, . . .) = (0, 0, 0, 0, . . .). Then it must be that c1 + c2 (0) = 0 ⇒ c1 = 0 (first element), c1 + c2 (1) = 0 ⇒ c2 = 0 (second element). This shows that B = {(1, 1, 1, 1, . . .), (0, 1, 2, 3, . . .)} is a basis for W . 2. Show that the linear transformation T (f (t)) = f (2t) from P2 to P2 (that is T (a + bt + ct2 ) = a + 2bt + 4ct4 ) is an isomorphism. Since we are given that T is linear, we only have to show that T is one to one and onto. To show that T is onto, we show that the equation T (f ) = a + b + ct2 has a solution f ∈ P2 for any choice of a, b, and c. If we take c b f (t) = a + t + t2 , 2 4 then T (f ) = a + bt + ct2, so T is onto. To show that T is one to one, we can just show that ker(T ) = {0}. Suppose that 0 = T (a + bt + ct2 ) = a + 2bt + 4ct2 . Then it must be true that a = b = c = 0 (by the fundamental theorem of algebra). This shows that T is also one to one, so it is an isomorphism.