Let $T$ be a linear operator on the vector space $V$ with $\operatorname{dim}(V)=2$ such that $T$ is not multiplication by a scalar. We have to show there is $v \in V$ such that $\{v, T(v)\}$ is a basis of $V$.


In particular, $T$ is not zero operator. Hence there is a $w \in V-\{0\}$ such that $T(w) \neq 0$. Now if $\{w, T(w)\}$ is linearly independent then we are done. Next consider the case when $\{w, T(w)\}$ is linearly dependent i.e. we have two scalars $\lambda$ and $\mu$, not both zero such that $\lambda w+\mu T(w)=0$. Actually $\lambda \neq 0$ and $\mu \neq 0$ as $T(w) \neq 0$ and $w \neq 0$. Choose $u \in V$ such that $\{u, w\}$ is a basis of $V$. Write $T(u)=\delta w+\alpha u$ for some scalars $\delta$ and $\alpha$. Let $v=\gamma w+u$ where $\gamma$ is a scalar such that $\gamma=0$ if $\delta \neq 0$ and $\gamma \neq 0$ if $\delta=0$. Note that $\delta=0$ implies $\alpha+\lambda \mu^{-1} \neq 0$ as $T$ is not multiplication by a fixed scalar. Now let $c_{1}$ and $c_{2}$ are two scalars, not both zero such that $c_{1} v+c_{2} T(v)=0$ i.e. $c_{1}(\gamma w+u)+c_{2}(\gamma T(w)+\delta w+\alpha u)=0$ i.e. $c_{1}(\gamma w+u)+c_{2}\left(-\gamma \lambda \mu^{-1} w+\right.$ $\delta w+\alpha u)=0$ i.e. $\left(c_{1}+c_{2} \alpha\right) u+\left(c_{1} \gamma-c_{2} \gamma \lambda \mu^{-1}+c_{2} \delta\right) w=0$. Now, $\{u, w\}$ is linearly independent implies $c_{1}+c_{2} \alpha=0=c_{1} \gamma-c_{2} \gamma \lambda \mu^{-1}+c_{2} \delta$. Therefore, $c_{2}\left(-\alpha \gamma-\gamma \lambda \mu^{-1}+\delta\right)=0$. Now $c_{2} \neq 0$ as $v \neq 0$. That is $\gamma\left(\alpha+\lambda \mu^{-1}\right)=-\delta$. By our choice of $\gamma$ this leads to a contradiction. Therefore, $\{v, T(v)\}$ is a linearly independent set. Hence we are done.

Please cheak this. Also you can give another approach. Thank you.

  • 1
    $\begingroup$ You can read the corresponding sections in Hoffman-Kunze. Such a vector is called a cyclic vector of $T$. A cyclic vector exists if and only if the minimal polynomial is the same as its characteristic polynomial. $\endgroup$
    – Q-Zhang
    May 6, 2022 at 2:25
  • 2
    $\begingroup$ In dimension $2$, the claim that $\{v, Tv\}$ is a basis is equivalent to $\{v,Tv\}$ being linearly independent. And $\{v,Tv\}$ is linearly independent if and only if $Tv \not\in \operatorname{span} \{ v\}$. So you would just need to show that there exists $v \in V$ such that $Tv \neq \lambda v$. Try to prove the following: if $Tv \in \operatorname{span} \{ v\}$ for all $v \in V$, then $T$ is multiplication by a scalar. $\endgroup$
    – spin
    May 6, 2022 at 2:28

1 Answer 1


Suppose such a pair does not exist. Then $\{v, Tv\}$ are linearly dependent for all $v \in V$. I.e., $Tv = k_v \cdot v$. Let $u, v \in V$ be two linearly independent vectors (hence a basis for $V$), with $Tv = k \cdot v,\ Tu = c \cdot u,\ T(u + v) = \lambda \cdot (u + v)$, and where $k = k(v), c = c(u), \lambda = \lambda(u + v)$. Then,

$$ \lambda u + \lambda v = \lambda (u + v) = T(u+v) = Tu + Tv = cu + kv \iff\\ (c - \lambda) u + (k - \lambda) v = 0 \iff\\ c = \lambda = k $$

and $T$ is multiplication by a scalar.


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