TITLE: Equivalent condition for a vector space to be finite dimensional
QUESTION [5 upvotes]: Let $k$ be a field and let $V$ be a vector space over $k$. Then $V$ is finite dimensional if and only if for every $\phi\in End_k(V)$, there are $a_0,\dots,a_{m-1}\in k$ such that 
  $$\phi^m+a_{m-1}\phi^{m-1}+\cdots+a_1\phi+a_0id_V=0.$$

I have no idea on how to prove this statement. I was trying to use the fact that $V$ is finite dimensional if and only if $End_k(V)$ is finite dimensional... Could you help me with this? Thanks!

REPLY [5 votes]: Suppose $V$ is finite dimensional, then so is $\mathrm{End}_k(V)$. In particular for any $\varphi \in \mathrm{End}_k(V)$ there exists some $n$ such that the set $\{\varphi^0,\dots,\varphi^n\}$ is linearly dependent. This $n$ is of course $\dim_k \mathrm{End}_k(V)$. Existence of such $a_i$ follows. 
I don't see an easy way to do the reverse direction using the fact you like. I'd rather prove it by contraposition. If $V$ is infinite dimensional take a countable linearly independent set of unit vectors $\{e_n\}$ and define $\varphi(e_n)=e_{n+1}$ and extend $\varphi$ linearly to the rest of the space. Then we have that $\varphi^k(e_1)=e_{k+1}$. So no such set of coefficients can exist, because the set $\{\varphi^k(e_1)\}_{k=0}^\infty$ is linearly independent.