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Definition: If $\alpha \in V$, the $T$-cyclic subspace generated by $\alpha$ is the subspace $Z(\alpha; T)=\{g(T)(\alpha)\mid g\in F[x]\}$. If $Z(\alpha; T)=V$, then $\alpha$ is called a cyclic vector for $T$.

The space $Z(\alpha; T)$ is one-dimensional if and only if $\alpha$ is a characteristic vector for $T$.

My attempt: It’s easy to check $Z(\alpha ;T)=\text{span}(\{T^k(\alpha)\mid k\geq 0\})$. $(\Rightarrow)$ Suppose $\dim Z(\alpha;T)=1$. Then $\exists \beta\in V$ such that $\text{span}(\beta)= \text{span}(\{T^k(\alpha)\mid k\geq 0\}) $. So $\alpha =c\cdot \beta$ and $T(\alpha)=d\cdot \beta$. Since $\alpha \neq 0$, we have $c\neq 0$. So $\exists c^{-1}\in F$ such that $cc^{-1}=c^{-1}c=1$. Thus $\beta =c^{-1}\cdot \alpha$ and $T(\alpha)=d\cdot \beta=(dc^{-1})\cdot \alpha$. Hence $\alpha$ is eigenvector of $T$.

$(\Leftarrow)$ Suppose $\alpha$ is eigenvector of $T$. Then $\exists c\in F$ such that $T(\alpha)=c\cdot \alpha$. So $T^k(\alpha)=c^k\cdot \alpha$, $\forall k\geq 0$. Thus $$Z(\alpha;T)=\text{span}(\{T^k(\alpha)\mid k\geq 0\})=\text{span}(\{c^k\cdot \alpha\mid k\geq 0\})=\text{span}(\alpha).$$ Since $\alpha \neq 0$, we have $\{\alpha\}$ is independent. So $\{\alpha\}$ is basis of $Z(\alpha ;T)$. Hence $\dim Z(\alpha ;T)=1$. Is my proof correct?

user264745
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    Filling in some details: (1) if $\alpha =0$, then $Z(\alpha ; T)={0}$. So $\dim Z(\alpha ; T)=0$. Thus we reach contradiction. (2) $\text{span}({c^k\cdot \alpha\mid k\geq 0})=\text{span}(\alpha)$. $\supseteq $ is trivial. Let $\beta \in \text{span}({c^k\cdot \alpha\mid k\geq 0})$. Then $\beta =\sum_{i=1}^n a_i\cdot (c^{r_i}\cdot \alpha)$. So $\sum_{i=1}^n a_i\cdot (c^{r_i}\cdot \alpha)= \sum_{i=1}^n (a_i\cdot c^{r_i})\cdot \alpha=\left( \sum_{i=1}^n a_i\cdot c^{r_i}\right)\cdot \alpha \in \text{span}(\alpha)$. Thus $\subseteq$. Hence equality. – user264745 Feb 04 '23 at 08:30
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    For (2) I would simply say $\text{span}({c^k\cdot \alpha\mid k\geq 0})\subset\text{span}(\text{span}(\alpha))=\text{span}(\alpha)).$ – Anne Bauval Feb 04 '23 at 09:00

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