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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](http://goo.gl/mLWc8), so it may attract downvotes, or closed. To prevent that, please [edit] the question. [This](http://goo.gl/PlJyVQ) 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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