5. Consider the following statement, which is true: "Let V be a vector space over a field F, and let Te S(V). Suppose that for some x e V, we have 73x = 0, and V = C(x), where C(x) is the cyclic subspace generated by x. Then 1 is not an eigenvalue of T." Read the following "proof of the statement, which is incorrect. "Proof": Suppose Tv = v for some ve V. We will show v = 0 and therefore 1 is not an eigenvalue of T. We are given V = C(x) = span{x, Tx, 72x, ...} and 73x = 0. Since 73x = 0, then T*x = 0 for all k 2 3, and V = span {x, Tx, T2x}. Therefore, there exist scalars co, C1, C2 E F such that v = cox + qTx+ c272x (1) Applying T to both sides of this expression, together with our assumption that Tv = v, gives v = coTx + al'x (2) Equating the expressions (1) and (2) for v, and rearranging gives Cox + (CI - co)Tx + (c2 - c)T2 x = 0 which implies co = 0, c1 = co = 0, and c2 = c1 = 0. Therefore, v = cox + ciTx + c272x = 0 which is what we wanted to show. What is wrong with this "proof"? Clearly identify any incorrect assumptions or conclusions in the "proof", and briefly explain why they are incorrect. Your answer should be no longer than one paragraph. [4 marks]