Prove that if \(T^2\) has a cyclic vector, then \(T\) has a cyclic vector. Is the converse true?
Solution
1. Let \(v\) be a cyclic vector for \(T^2\), so \(\{v, T^2 v, T^4 v, \ldots\}\) spans \(V\).
definition of cyclic vector.
2. The cyclic subspace generated by \(v\) under \(T\) is \(Z(v, T) = \operatorname{span}\{v, Tv, T^2 v, \ldots\}\).
definition.
3. \(Z(v, T)\) contains \(\{v, T^2 v, T^4 v, \ldots\}\) (the even powers), which span \(V\) by
1.
1 and 2.
4. Hence \(Z(v, T) = V\), so \(v\) is a cyclic vector for \(T\).
3.
5. The converse is false.
answer to the second question.
6. Counterexample: let \(T\) be the operator on \(\mathbb{R}^2\) given by \(T(x, y) = (y, 0)\) (a nilpotent operator with \(T^2 = 0\)).
choose a counterexample.
7. \(T\) has a cyclic vector (e.g. \(v = (1, 0)\), since \(\{v, Tv\} = \{(1,0), (0,1)\}\) spans \(\mathbb{R}^2\)), but \(T^2 = 0\) has no cyclic vector (its cyclic subspaces are all \(1\)-dimensional).
6.
8. Hence the converse is false.
7.
9. Q.E.D.
4 and 8.