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5. Consider the following statement, which is true: \"Let V be a vector space over a eld F, and let T E (V). Suppose that for some 3; E V, we have T3); = 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 v E V. We will show v = 0 and therefore 1 is not an eigenvalue of T. We are given V = C(x) = span{x,Tx,T2x,...} and T3): = 0. Since T3): = 0, then Tkx = 0 for all k 2 3, and V = span{x, TX, T234}. Therefore, there exist scalars 60,61,62 E F such that v = cox + clTx + C2T2x (1) Applying T to both sides of this expression, together with our assumption that Tv = v, gives v = CDTX+ c1T2x (2) Equating the expressions (1) and (2) for v, and rearranging gives cox + (cl CD)TX + (62 c1)T2x = 0 which implies c0 = 0, c1 = :30 = 0, and c2 = c1 = 0. Therefore, v = cox + clTx + c2T2x = 0 which is what we wanted to show. What is wrong with this \"proof\"? Clearly identify any incorrect assumptions or conclusions in the \"proof\