TITLE: Artin's Algebra 4.4.8
QUESTION [5 upvotes]: Q.
Let $T$ be a linear operator on a finite-dimensional vector space for which every nonzero vector is an eigenvector. Prove that $T$ is multiplication by a scalar.
I did not find this question in old posts.
Approach:-
First suppose $\operatorname{dim}(V)=1$. Then there is a non-zero vector $v \in V$ such that $V=\{c v: c \in F\}$. By hypothesis we have $\lambda \in F$ such that $T(v)=\lambda v$, so that $T(c v)=c T(v)=c(\lambda v)=\lambda(c v)$ i.e. $T=\lambda I$ where $I: V \rightarrow V$ is the identity operator. So in this case we are done.
Next suppose, $\operatorname{dim}(V) \geq 2$. Then let $u, w$ be two linearly independent vectors of $V$. Now we have $\alpha, \beta \in F$ such that $T(u)=\alpha u$ and $T(w)=\beta w$. Now note that $u+w \neq 0$ as $\{u, w\}$ is a linearly independent set. Hence there is $\gamma \in F$ such that $T(u+w)=\gamma(u+w)$. So that $\alpha u+\beta w=T(u)+T(w)=T(u+$ $w)=\gamma(u+w)$. Hence $\alpha u+\beta w=\gamma u+\gamma w$ i.e. $(\alpha-\gamma) u=(\gamma-\beta) w$. Since $\{u, w\}$ is a linearly independent set we have $\alpha-\gamma=0=\gamma-\beta$ i.e. $\alpha=\beta$. What we observe is that for every vector $w$ which is linearly independent with $u$ we have $T(w)=\alpha u$ where $\alpha \in F$ is such that $T(u)=\alpha u$. Now every linearly independent subset can be extended to a basis of $V$. So let $\left\{v_{1}, \ldots, v_{n}\right\}$ be a basis of $V$ with $u=v_{1}$, then for any $x \in V$ with representation $x=c_{1} v_{1}+c_{2} v_{2}+\ldots+c_{n} v_{n}$ with $c_{1}, \ldots, c_{n} \in F$ we have $T(x)=c_{1} T\left(v_{1}\right)+$ $c_{2} T\left(v_{2}\right)+\ldots+c_{n} T\left(v_{n}\right)=c_{1}\left(\alpha v_{1}\right)+c_{2}\left(\alpha v_{2}\right)+\ldots+\left(c_{n} \alpha v_{n}\right)=\alpha\left(c_{1} v_{1}+\ldots+c_{n} v_{n}\right)=\alpha x$ i.e. $T=\alpha I$.
The case when $\operatorname{dim}(V)=0$ is trivial as in this case $V=\{0\}$ so that $T=0=0 I$, where $I: V \rightarrow V$ is the identity operator.
Please cheak this.. Also you can give your approach. Thank you...

REPLY [3 votes]: This has certainly been proved on MSE before.
Pick some $v \neq 0$ then $Av = \lambda v$ (the same relationship holds for any scalar multiple of $v$, of course).
Pick any $u$ that is not colinear with $v$ then
$A(u+v) = \alpha(u+v)$ and $A u = \beta u$ for some $\alpha, \beta$.
Since $\alpha u + \alpha v = \lambda v + \beta u$, linear independence shows that $\alpha = \lambda = \beta$ and so $Au = \lambda u$.
Hence $A = \lambda I$.