# Artin's Algebra $4.3.3$

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 $$V$$.

Approach:-

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.

• 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. May 6, 2022 at 2:25
• 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.
– spin
May 6, 2022 at 2:28

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.