Question: Suppose V is a finite-dimensional vector space and I have a linear subspace of its endomorphisms
$$W \subseteq \mbox{End}(V).$$
How can I easily check if every vector of $V$ is fixed by some element of $W$? I would also be interested in any nice conditions on $W$ that imply 
$$\forall v \in V: \;\;\; v \in Wv,$$
even if they aren't biconditional.

* If the identity matrix happens to be in $W$, then the condition is satisfied trivially. However, it is easy to find examples where W omits the identity but can still fix any vector. For example, the two-by-two traceless matrices have this property.
* Unfortunately, $W$ can be large without satisfying the property. For example, take all matrices which have a zero in the upper left. These can never fix the first basis vector.

---

Side note:

In my particular situation, $W$ happens to be closed under multiplication. If this turns out to be a helpful condition, please don't hesitate to use it.

Answer: 
The question is closely related to the notion of algebraic reflexivity. 
The (algebraic) reflexive closure of $W$ is the space of all operators $g \in End(V)$ such that $g(x) \in Wx$ for all $x \in V$. 
The author asks when the identity operator belongs to the reflexive
closure of the operator space $W$. 

I will consider $W$ as a matrix space to simplify things. 
By a lemma of Azoff ([On Finite Rank Operators and Preannihilators](http://books.google.fr/books?id=S8gHeZah-2cC&printsec=frontcover&hl=fr&source=gbs_ge_summary_r&cad=0#v=onepage&q&f=false)), the reflexive closure of $W$ is the orthogonal
of the span of the rank $1$ matrices in $W^\bot$ (the orthogonal of $W$ for the 
trace bilinear form $(A,B) \mapsto tr(AB)$). 
Thus, the identity is in the reflexive closure of $W$ if and only if it is orthogonal 
to all the rank $1$ matrices in $W^\bot$. Put differently, 
every vector of $V$ is fixed by at least one operator in $W$ if and only if all the rank $1$ operators in $W^\bot$ have trace zero.

Using this insight, one recovers the case of hyperplanes that was explained earlier by Harry Altman. 

Now, if we add the condition that $W$ be stable under multiplication and that the underlying field $F$ be algebraically closed, then the identity belongs to the reflexive closure of $W$ if and only if it belongs to $W$!
This combines ideas from other answers: if $W$ does not contain the identity, it contains
only singular operators. One can use the Burnside theorem as it actually applies to 
every linear subspace that is stable under multiplication (provided that the space $V$ has dimension larger than $1$), not only to those which contain the identity. Using this repeatedly, one finds a basis of $V$ in which $W$ is represented by a space of block-upper-triangular matrices, where each diagonal block space
is either a full matrix space $M\_p(F)$ or $\{0\}$ (with the latter consisting of the zero $1 \times 1$ matrix). As the underlying field $F$ is infinite, one sees that at least one of those diagonal block spaces must be $\{0\}$ (otherwise, one shows that $W$ contains an invertible operator). With such a block space, the corresponding vector $x$ of $V$ is fixed by no operator in $W$. 

With a similar method, but relying on Wedderburn's structure theorem for irreductible
subalgebras instead of Burnside's theorem (which is just a special case of it), 
one can extend the above result to the more general situation where $W$ is stable under multiplication and the underlying field has cardinality greater than or equal to 
$\dim V$.
I'm going to add another partial answer, this time regarding the case where we do require that $W$ be closed under multiplication.

Namely: If $W$ is closed under multiplication, and $\dim W>n^2-2n+2$, then the only way $W$ can fix every vector is if $W$ contains the identity.

As Julian Rosen pointed out in the comments, if $W$ is closed under multiplication and contains any invertible element, it then contains the identity. So in order for $W$ to fail to fix some vector, it must have no invertible elements.

Fortunately, large subspaces of matrices containing no invertible elements (more generally, having elements of bounded rank) have been much studied! Here I'll use the results of ["The classification of large spaces of matrices with bounded rank"](http://arxiv.org/abs/1004.0298), by Clément de Seguins Pazzis. This will get us a converse to the above statement. For according to this, if $\dim W>n^2-2n+2$ and $W$ has no invertible elements, then either there exists a nonzero $v\in V$ which is in the kernel of all elements of $W$ (and so $W$ cannot fix every vector), or there is some $n-1$-dimensional subspace of $V$ which contains the images of all elements of $W$ (and so $W$ cannot fix every vector).

Thus, for $\dim W> n^2-2n+2$, being able to fix every vector is equivalent to containing some invertible element, which is equivalent to containing the identity.

(Note, by the way, that if $\dim W > n^2-n$, then automatically $W$ must contain some invertible element.)

Now the paper also proves results about what happens when $\dim W=n^2-2n+2$, but I don't see how to apply these, because while the resulting subspaces either A. don't fix every vector or B. aren't closed under multiplication, I don't think these properties are preserved under the notion of equivalence used in the paper, and so it's not so clear what happens then. Still, I think one ought to be able to do better than what I've written here -- we've only used closure under multiplication to go from "$W$ contains some invertible element" to $I \in W$, and I think one ought to be able to do much more with it than that.