Term for "pre-image, but in a vector-space-y way"? Consider a linear transformation $T :: V \rightarrow W$. Today I found myself wanting to use "the pre-image of $T$" to talk about "the subspace of $V$ that doesn't get killed by $T$".
I guess that set would be defined as $V - \ker(V) \cup \{0\}$; I took away the kernel, which contains zero, so I need to add back the zero to make it a subspace.
Is there a better term for this? It seems to me like this is solving a similar problem to the one direct sum solves -- we'd like to say "it's the sum of two disjoint subspaces", but they're not literally disjoint because their intersection is $\{0\}$. So we introduce a new concept.
 A: There is no term for such a subspace because there is no such subspace. The sum of two vectors that do not get killed, can get killed.  
As in your example let T be the projection $R^2\rightarrow R$ of the $xy$-plane onto the $x$-coordinate on the line.  The subset you describe contains all vectors $\langle x,y\rangle$ with $x\neq 0$.  This is nearly all of the plane. It is all except for the $x$-axis.  This subset, though it is not a vector subspace, can just be called the origin plus the complement of the $x$-axis as you suggest.
There is a term for a related vector space though: the image $\mathrm{Im}(T)$, which is $V$ with the kernel modded out.  $V/\mathrm{ker}(T)$. (In an earlier version I misspoke and called this the cokernel of $T$.  It is the cokernel of the kernel, but not the cokernel of $T$.) This is not a subspace of $V$ but in fact an image of $V$, specifically isomorphic to the image of $V$ under the map $T$ (a subspace of $W$).  I can explain further if needed.  It is an easy Google search.
Given an inner product on $V$ there is another well defined vector subspace of $V$, namely the space orthogonal to the kernel.  There are various notations for this, but no name in words shorter than the orthogonal subspace to the kernel.  Different choices of inner product will give different orthogonal subspaces.  In any case (unless the inner product is trivially degenerate) this subspace will not include all the vectors that are not killed by $T$. 
