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I saw this on A. I. Kostrikin's Introduction to Algebra Sect.2.3 Exercises.

If $\mathcal{A}\in \text{Hom}(V,V)$ in which $\dim V=n$ is such that $\mathcal{E}=\text{id}_V,\mathcal{A},\mathcal{A}^2,\cdots,\mathcal{A}^{n-1}$ is linearly independent in Hom$(V,V).$ Show that V is cyclic, i.e., there is some $v\in V$such that $$V=\text{span}\{v,\mathcal{A}v,\cdots,\mathcal{A}^{n-1}v\}.$$

Proofs using Hamilton-Cayley's theorem or not are all welcome.

Edit: I appreciate the kind administrators of MSE who gave me advice on how to modify the problems so that better answers can be given.

Now I give my thoughts on the problem. First If $\text{Im} \mathcal A^{p}=\text{Im} \mathcal A^{p+1}$for some $p$,then by a well-known fact $V=\text{Ker} \mathcal A^p\oplus \text{Im}\mathcal A^p,$which reminds me of mathematical induction in which this equation can be used to reduce dimensions.But then I'm at a loss and have no idea how to prove the theorem.

Then I used Hamilton-Cayley's theorem, but that doesn't seem to make things better.

Ramos Liang
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    Welcome to MSE. Your question is phrased as an isolated problem, without any further information or context. This does not match many users' quality standards, so it may attract downvotes, or closed. To prevent that, please [edit] the question. This will help you recognise and resolve the issues. Concretely: please provide context, and include your work and thoughts on the problem. These changes can help in formulating more appropriate answers. – José Carlos Santos Jun 12 '22 at 07:41
  • @JoséCarlosSantos Thank you, I've edited my question. – Ramos Liang Jun 12 '22 at 07:54
  • I guess from the context that $\dim V=n.$ – Ryszard Szwarc Jun 12 '22 at 08:08
  • @RyszardSzwarc Yes, I forgot to mention. – Ramos Liang Jun 12 '22 at 08:13
  • Are you not familiar with the result that a linearly independent set of cardinality equal to the dimension is automatically a basis for the space? – Mr.Gandalf Sauron Jun 12 '22 at 08:17
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    @Mr.GandalfSauron I'm sorry, but I AM familiar with the elementary result that you are mentioning. That is the first proposition I met when studying linear algebra. However, it is not given that the vectors above are linearly independent. What is actually given is linear independence in Hom(V,V). – Ramos Liang Jun 12 '22 at 08:25

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