TITLE: Spectrum of a linear operator on a vector space of countable dim
QUESTION [11 upvotes]: How would you prove that if V is a vector space over $\mathbb{C}$ of countably infinite dimension, and $T$ is a linear operator on V, then Spectrum($T$) is non-empty?

REPLY [6 votes]: This is a comment to Rankeya's answer posted as a community wiki answer. 
Dear Rankeya, your answer looks correct to me, but I think the writing could be improved. Here is how I would have written it. I don't claim that my own writing is good. I'm just trying to help. Dear users, you're all welcome to suggest a better wording. 
Suppose by contradiction there is a linear operator $T$ on $V$ such that 
(1) $T-\lambda$ is an isomorphism for all $\lambda\in\mathbb C$. 
Claim. There is a (unique) $\mathbb C$-algebra morphism $\phi$ from $\mathbb C(X)$ to $\text{End}(V)$ which maps $X$ to $T$. (Here $X$ is an indeterminate.) 
Proof of the Claim. Let $p\in\mathbb C[X]$ be nonzero. It suffices to check that $p(T)$ is invertible. But this follows from (1). QED 
The formula $fv:=\phi(f)v$, for $f\in\mathbb C[X]$ and $v\in V$, turns $V$ into a nonzero $\mathbb C(X)$-vector space. To get the sought-for contradiction, it suffices to prove that the dimension of $\mathbb C(X)$ over $\mathbb C$ is uncountable. But otherwise the set of all $(X-\lambda)^{-1}$, where $\lambda$ runs over $\mathbb  C$, would be $\mathbb C$-linearly dependent, contradicting the uniqueness of the partial fraction decomposition.