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Let $T$ be a linear operator on $F^2$. Prove that any non-zero vector which is not a characteristic vector for $T$ is a cyclic vector for $T$. Hence, prove that either $T$ has a cyclic vector or $T$ is a scalar multiple of the identity operator.

My attempt: let $\alpha \in F^2\setminus \{0\}$ and $\alpha$ is not eigenvector of $T$. By example 1 section 7.1, $\dim Z(\alpha;T)\neq 1$. Since $\alpha \in Z(\alpha; T)$, we have $Z(\alpha;T)\neq 0$. So $\dim Z(\alpha;T)\neq 0$. Thus $\dim Z(\alpha;T)=2$. Hence $Z(\alpha ;T)=F^2$.

If $\exists \alpha\in F^2\setminus \{0\}$ such that $\alpha$ is not eigenvector of $T$, then $T$ has a cyclic vector, namely $\alpha$. Suppose $\alpha$ is eigenvector of $T$, $\forall \alpha \in F^2\setminus \{0\}$. Then $T(\alpha)=c_{\alpha}\cdot \alpha$, for all $\alpha \in F^2\setminus \{0\}$. Let $\alpha ,\beta \in F^2\setminus \{0\}$. Case (1): $\{\alpha ,\beta\}$ is independent. Then $T(\alpha +\beta)=c_{\alpha +\beta}\cdot (\alpha +\beta)=c_{\alpha +\beta}\cdot \alpha +c_{\alpha+\beta}\cdot \beta=c_\alpha \cdot \alpha +c_\beta \cdot \beta$. So $c_\alpha=c_\beta=c_{\alpha +\beta}$. Case (2): $\{\alpha ,\beta\}$ is dependent, i.e. $\alpha =c\cdot \beta$, for some $c\in F$. Then $T(\alpha)=c\cdot (c_\beta \cdot \beta)=c_\beta \cdot (c\cdot \beta)=c_\beta \cdot \alpha=c_\alpha \cdot \alpha$. So $c_\alpha =c_\beta$. Thus $c_\alpha =c_\beta$, $\forall \alpha,\beta \in F^2\setminus \{0\}$. Hence $T$ is scalar multiple of identity operator. Is my proof correct?

user264745
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  • If $\alpha$ is an eigenvector of $T$, then $T(\alpha)=c_{\alpha}\alpha$ for that particular $\alpha$, not for all $\alpha$. – C Squared Feb 07 '23 at 21:31
  • First note that $\dim Z(\alpha,T)=0$ if and only if $\alpha=0$ (If $\alpha=0$, then this is clear. If $\alpha\neq 0$, then $\alpha\in Z(\alpha, T)$, hence $\dim Z(\alpha,T)\neq 0$).

    See if this question helps: If $T$ does not have a cyclic vector, then what is the dimension of $Z(\alpha, T)$ for non-zero $\alpha\in V$?

     – C Squared Feb 07 '23 at 21:45
  • @CSquared there is abuse of notation. $0\lt \dim Z(\alpha ;T)\lt \dim (V)$. – user264745 Feb 07 '23 at 22:09
  • I’m not sure what you mean by this… – C Squared Feb 08 '23 at 09:45
  • @CSquared first sentence answer your first comment and second sentence answer “see if this questions helps….”. I used $\alpha$ in two places: (1) $T(\alpha)=c_\alpha \cdot \alpha$, for all $\alpha \in F^2\setminus {0}$ (2) let $\alpha ,\beta\in F^2\setminus {0}$. So there might be some ambiguity for use of $\alpha$. – user264745 Feb 08 '23 at 11:40
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    okay, I think that clears things up then. the proof looks good. my previous hint was trying to get at the fact that if there is no cyclic vector, then every non-zero vector is an eigenvector in this case – C Squared Feb 08 '23 at 17:40

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