TITLE: Subspaces of End(V) that can fix any vector
QUESTION [8 upvotes]: 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.

REPLY [5 votes]: For subspaces that are stable under multiplication, there's an even more profound
result that holds for all fields : 

Given a finite-dimensional vector space $V$ and a linear subspace $W$ of $End(V)$
  that is stable under multiplication but does not contain the identity, 
  there are linear subspaces $V_1$ and $V_2$ of $V$ such that $V_1$ is a proper subspace of
  $V_2$ and every operator in $W$ maps $V_2$ into $V_1$. 
  In particular, every operator in the reflexive closure of $W$ must map $V_2$ into $V_1$, 
  whence the reflexive closure of $W$ consists only of singular operators!

The proof is essentially similar to the one I have outlined in my first answer. 
The key is a theorem of Wedderburn which states that a subalgebra of $End(V)$
is either reducible or simple. If we have a linear subspace $W$ which is stable under multiplication, then we note that $W +Span(id)$ is a subalgebra of $End(V)$ and $W$
is a two-sided ideal of it: thus, either $W$ is irreducible, or $W$ is a simple algebra,
or $W$ is limited to the zero operator and $V$ has dimension $1$. 
Using this repeatedly, one finds a basis in which all the operators in $W$ are represented by block-upper-triangular matrices, and, on the diagonal, each block space is an algebra 
or consists of the $1 \times 1$ zero matrix. 
If there is at least one block of the latter time, then we have found the subspaces $V_1$ an $V_2$ we were looking for. 
If we now assume that all the diagonal block spaces
are algebras, then we prove that $W$ contains a non-singular matrix, which yields, 
as explained in some answers, that $W$ contains the identity. 
To do this, it suffices to prove the following lemma :

Let $A_1,\dots,A_n$ be algebras, and $H$ be a linear subspace of 
  $A_1\times \cdots \times A_n$ that is stable under multiplication and projects surjectively
  onto each $A_i$. Then, $H$ contains the unit element of $A_1\times \cdots \times A_n$. 

This is proved by induction over n. The case $n=1$ is trivial. Assume that this holds for
$n-1$, with $n \geq 2$ fixed. Then, we write 
$A_1\times \cdots \times A_{n-1}=B$; by induction, $H$ contains $(1_B,x)$ for some
$x \in A_n$. Then, as some $b \in B$ satisfies $(b,1_{A_n}) \in B \times A$, we see that
$(1_B,1_{A_n})=(1_B,x)+(b,1_{A_n})-(1_B,x)*(b,1_{A_n}) \in H$.