If linear operator $T^2 $has a cyclic vector then so does $T$ Here is my reasoning: 
$T^2$ has a cyclic vector iff its minimal polynomial and its characteristic polynomial are equal.
Minimal polynomial of $T^2$ has degree $n$ . Therefore $T$ has an annihilating polynomial with degree $2n$ and with variables having even powers.
Since minimal polynomial of $T$ divides the annihilating polynomial having even powers, the powers of $x$ in minimal polynomial of $T$ will have the same parity. If minimal polynomial of $T$ is $q(x)$ and deg q(x) < n , $q(x)q(x)=p'(x^2)$ and p' is an annihilating polynomial of $T^2$ having deg < n which is a contradiction.
Is my reasoning is correct?

I don't understand why go through all the trouble. Vector $x$ is cyclic for $T$ if vectors $T^nx$ span whole space. Clearly, vectors of the form $T^{2n}x$ are a subset of these.
