# Artin's Algebra 4.4.8 [duplicate]

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.

• This looks correct to me.
– Koro
May 6, 2022 at 3:50
• Isn’t exercise 4.4.8 just a little iteration on exercise 4.3.3? May 7, 2022 at 4:04
• @JoeShmo I can't see how. Sep 11, 2022 at 12:12
• Does this answer your question? Eigenvectors of a Homothety operator Sep 11, 2022 at 12:21
• @AnneBauval the solution that OP is outlining in his very question (is the solution that copper.hat gave him) is the solution that I gave him for exercise 4.3.3 earlier that day. Jun 16 at 18:54

We can do this without considering the number of dimensions separately.

Let $$V$$ be the vector space and let $$v\in V$$. Since $$V$$ is finite dimensional we have a finite basis $$\{e_1,e_2,...,e_n\}$$ where $$n$$ is the dimension of $$V$$.

We write $$v=\Sigma_{i=1}^{n}a_ie_i \text{ and since }v \text{ is an eigenvector, }T(v)=\lambda v=\Sigma_{i=1}^{n}\lambda a_ie_i=\Sigma_{i=1}^{n}a_iT(e_i)$$

We can see that $$T(e_i)=\lambda e_i$$

Since $$\lambda$$ is the same eigenvalue for each of the basis vectors, we get that for any $$v=a_ie_i$$, $$T(v)=\lambda a_i e_i=\lambda T(v)$$ which proves that $$T$$ is multiplication by $$\lambda$$.

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 $$A = \lambda I$$.