Is a virtual vector bundle the same as a vectorial bundle? What is a virtual vector bundle?
Is a virtual vector bundle the same as a vectorial bundle?
The current entry in nLab states the following:

"In one class of models for K-theory – generalized (Eilenberg-Steenrod) cohomology theory – cocycles are represented by Z2-graded vector bundles (pairs of vector bundles, essentially) modulo a certain equivalence relation. In that context it is sometimes useful to consider a certain variant of infinite-dimensional Z2-graded vector bundles called vectorial bundles.
Much else to be discussed..."
Source: http://ncatlab.org/nlab/show/vector+bundle#virtual_vector_bundles

The ellipses, rather than answering my question, have led me to wonder.
 A: A virtual vector bundle can be thought of as being either a vector bundle or the "opposite" of a vector bundle. This construction takes advantage of the fact that, under direct sum, the set of isomorphism classes of vector bundle over a fixed base $X$ are essentially a monoid. In particular, we obtain a monoid structure if we identify vector bundles $E$ and $E'$ whenever there is are trivial bundles $\epsilon$ and $\epsilon'$ so that $E \oplus \epsilon$ and $E' \oplus \epsilon'$ are isomorphic. This is a monoid with identity $\epsilon$ corresponding to the class of trivial bundles. You can think of this collection of vector bundle classes as being similar to $\mathbb{N}$ under addition. 
The additive group $\mathbb{Z}$ can be constructed from $\mathbb{N}$ by taking equivalence classes of pairs $(a,b)$ under the relation that $(a,b) \sim (c,d)$ iff $a+d=b+c$. We add two pairs in a coordinate-wise fashion. This induces a well-defined operation on the set of equivalence classes of $(a,b)$. We can then identify $n$ with $[(n,0)]$ and define $-n = [(0, n)]$.
A virtual bundle is then a representative of the isomorphism classes of pairs $(E, E')$. Any vector bundle $E$ is then identified with $[(E, \epsilon) ]$. The inverse of $E$ is identified with $[( \epsilon ,E)]$.
This constructs an abelian group denoted $K(X)$. 
I was unfamiliar with vectorial bundles before tonight, but they are not the same. Apparently, there is an abelian group structure on these bundles which is isomorphic to $K(X)$, so classes of vectorial bundles are in bijective correspondence with virtual bundles.
