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Let T be a linear operator on a finite-dimensional vector space V.

(a) There exists a vector $\alpha$ in $V$ such that the $T$-annihilator of $\alpha$ is the minimal polynomial for $T$.

(b) $T$ has a cyclic vector if and only if the characteristic and minimal polynomials for $T$ are identical.

My attempt: (a) Let $W_0=\{0\}$. Then $W_0$ is proper $T$-admissible subspace of $V$. By theorem 3 section 7.2, $\exists \alpha_1,…,\alpha_r\in V\setminus \{0\}$ with respective $T$-annihilator $p_1,…,p_r$ such that $V=Z(\alpha_1;T)\oplus \cdots \oplus Z(\alpha_r;T)$ and $p_k|p_{k-1}$, for all $2\leq k\leq r$. Let $m$ be minimal polynomial of $T$. We claim $p_1=m$. Clearly $p_1$ is monic. Since $p_k|p_{k-1}$, $k=2,…,r$, we have $p_i|p_1$, $\forall i\in J_r$. So $\exists q_i\in F[x]$ such that $p_1=p_iq_i$. Let $\beta\in V$. Then $\exists !\ \beta_i\in Z(\alpha_i;T)$ such that $\beta =\beta_1+…+\beta_r$. By theorem 1 section 7.1, $p_i$ is minimal polynomial of $U_i=T|_{Z(\alpha_i;T)}$. Since $p_i(U_i)=0$, we have $p_i(T)(\beta_i)=0$. So $$\begin{align} p_1(T)(\beta)&= p_1(T)(\beta_1)+…+p_1(T)(\beta_r)\\&=(p_1q_1)(T)(\beta_1)+…+(p_rq_r)(T)(\beta_r)\\ &=q_1(T)p_1(T)(\beta_1)+…+q_r(T)p_r(T)(\beta_r)\\&= 0+…+0=0. \end{align}$$ Thus $p_1(T)(\beta)=0$, $\forall \beta \in V$. That is $p_1(T)=0$. Suppose $g\in F[x]$ such that $g(T)=0$. Then $g(U_i)=0$. So $\deg (p_i)\leq \deg (g)$, $\forall i\in J_r$. Thus $\deg (p_1)\leq \deg (g)$. By uniqueness, $p_1=m$. Hence $\exists \alpha_1\in V$ such that $T$-annihilator of $\alpha_1$ is the minimal polynomial of $T$.

(b) $(\Rightarrow)$ Since $T$ has a cyclic vector, we have $\exists \alpha \in V$ such that $Z(\alpha;T)=V$. Let $p_\alpha$ be $T$-annihilator of $\alpha$ and $m$ be minimal polynomial of $T$. By theorem 1 section 7.1, $\deg (p_\alpha)=\dim Z(\alpha;T)=n$ and $p_\alpha =m$. So $\deg (m)=n$. Let $f$ be characteristic polynomial of $T$. By this post, $f=x^n-\text{tr} (T)x^{n-1}+…+\det (T)$. So $f$ is monic and $\deg (f)=n$. By Cayley-Hamilton theorem, $f(T)=0$. By uniqueness, $f=m$.

$(\Leftarrow)$ Suppose $f=m$. By (a), $\exists \alpha \in V$ such that $p_\alpha =m$, where $p_\alpha$ is $T$-annihilator of $\alpha$. So $\deg (p_\alpha)=\deg (m)=\deg (f)=n$. By theorem 1 section 7.1, $\deg (p_\alpha)=\dim Z(\alpha ;T)=n$. Thus $Z(\alpha ;T)=V$. Hence $T$ has a cyclic vector. Is my proof correct?

Here is alternative proof of (a).

user264745
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  • What is a "T-annihilator of a vector"? Also "T-admissible subspace" ... I don't think these are general (international), well-known standard names so it'd be better to define them. – DonAntonio Feb 17 '23 at 12:42
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    @DonAntonio $T$-annihilator of $\alpha$ is monic generator of ideal ${g\in F[x]\mid g(T)(\alpha)=0}$. $W$ is $T$-admissible subspace of $V$ if $W$ is invariant under $T$ and if $f(T)(\beta)\in W$, then $\exists \gamma\in W$ such that $f(T)(\beta)=f(T)(\gamma)$. I thought these notions were obvious because it is fundamental to Cyclic Decomposition theorem. – user264745 Feb 17 '23 at 12:49

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