TITLE: Whether a given algebra is the algebra of endomorphisms for a vector space
QUESTION [7 upvotes]: Let $\mathbb{F}$ be a field and let $A$ be an associative unital $\mathbb{F}$-algebra. Is there a criterion to let me know if $A$ is isomorphic to the algebra $\mbox{End}(\mathbf{V})$ of endomorphisms for some $\mathbb{F}$-vector space $\mathbf{V}$? Generalizations to $A$ being an associative unital ring and $\mathbf{V}$ an Abelian group or similar are welcome. Answers for particular cases $\mathbb{F}\in\{\mathbb{R},\mathbb{C}\}$ are also appreciated. Thank you.
This is a repetition of the following question I posed on MathSE days ago which has no answers so far. As I am no expert in algebra, so I thought maybe it is a research level question.

REPLY [3 votes]: If $V$ is a vector space, the two-sided ideals of $E:=End(V)$ form a chain. The nonzero ideals are exactly those of the form $I_{\alpha}=\{\varphi\in E : {\rm dim}({\rm im}(\varphi))<\alpha \}$, where $\alpha$ is an infinite cardinal. 
Thus, for instance, if $V$ is a vector space of countable dimension, then there are three ideals in $E$.  Viewing $E$ as the ring of column-finite matrices, these ideals are exactly the zero ideal, the ideal consisting of matrices with only finitely many nonzero crows, and $E$ itself.
The one-sided ideals are much more complicated.  In particular, the Jacobson radical is the zero ideal.
The ring $E$ is von Neumann regular; meaning that for every $x\in E$ there exists some $y\in E$ with $xyx=x$.
The ring $E$ is clean, meaning that for every $x\in E$ there exists some unit $u$ and idempotent $e$ such that $x=e+u$.
There are many other purely ring theoretic properties that such endomorphism rings have.  A good source of information is Lam's "A First Course in Noncommutative Rings".
I know of no simple criterion for being an endomorphism ring of a vector space over a field.  (That's not to say there isn't one.)