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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...

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    $\begingroup$ This looks correct to me. $\endgroup$
    – Koro
    Commented May 6, 2022 at 3:50
  • $\begingroup$ Isn’t exercise 4.4.8 just a little iteration on exercise 4.3.3? $\endgroup$
    – Joe Shmo
    Commented May 7, 2022 at 4:04
  • $\begingroup$ @JoeShmo I can't see how. $\endgroup$ Commented Sep 11, 2022 at 12:12
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    $\begingroup$ Does this answer your question? Eigenvectors of a Homothety operator $\endgroup$ Commented Sep 11, 2022 at 12:21
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    $\begingroup$ @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. $\endgroup$
    – Joe Shmo
    Commented Jun 16, 2023 at 18:54

2 Answers 2

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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 $\lambda$.

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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 $A = \lambda I$.

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