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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!

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Are you sure your quantifiers are in the right order? You probably want $V$ is finite dimensional if and only if for every $\phi$ there exist $a_0,\ldots, a_{m-1}\in k$ such that $\phi^m + \cdots + a_0id_V = 0$. –  froggie Dec 15 '12 at 3:04
    
@froggie Thanks! You are absolutely correct. –  Vladimir Dec 15 '12 at 3:09

1 Answer 1

up vote 5 down vote accepted

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.

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Thanks! Clean proof. –  Vladimir Dec 15 '12 at 3:13
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Just a comment on "This $n$ is of course $\dim_k\operatorname{End}_k(V)$" (which is $(\dim_k V)^2$). While it is indeed most obvious that this $n$ suffices, it may be remarked that the in general much smaller value $\dim_k V$ will also do, for instance by the Cayley-Hamilton theorem. –  Marc van Leeuwen Dec 15 '12 at 9:43

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