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.




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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.



Subspaces of End(V) that can fix any vector
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 
$\dim V$. 


