# 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?

• A simpler argument is $V=Z(v,T^2) \subseteq Z(v,T)$ implies $V=Z(v,T)$ . – lhf Feb 21 '17 at 17:06
• @lhf : I saw this answer but I wanted to know if its possible to use the minimal polynomial argument. – A. Napster Feb 21 '17 at 17:08
• "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." Is this a well known statement about polynomials? – Omnomnomnom Feb 21 '17 at 18:18
• How can you guarantee that $q(x)q(x)$ will be of the form $p'(x^2)$? – Omnomnomnom Feb 21 '17 at 18:19
• Let's suppose for now that I accept your explanation for the first comment. I'm still skeptical about that second one. $q(x) = x^3 + 1$ leads to $q(x)q(x) = x^6 + 2x^3 + 1$, which is not a function of $x^2$. – Omnomnomnom Feb 21 '17 at 18:59

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.
• This is false. Take $T=\begin{pmatrix} 0 & 1 \\ 1 & 0 \end{pmatrix}$. Then $T$ has a cyclic vector $(1,0)^T$, but $T^2=1$ so it doesn't have a cyclic vector. – Blazej Jul 10 '19 at 7:14