My background is in physics. I am motivated to understand tensor behind "a tensor is something that transforms like a tensor." My understanding of a tensor in that a second rank tensor is an elements of $V\otimes V$. I very basic understanding of multilinear maps, but have no understanding of exterior products.
My initial assumption was that if a tensor, T, is an element of $V\otimes V$, there must exist some $\vec{v}, \vec{w} \in V s.t. T=\vec{v} \otimes \vec{w}$. This clearly cannot hold for antisymmetric tensors or a "null" tensor where all the elements are zero. I am not wondering how it is possible to construct an antisymmetric tensor (or, for that matter, other tensors with zero elements) from a set of vectors in V. If so how? (Someone suggested linear combination)?
Here is what I think you are asking: is every antisymmetric tensor of the form $v \wedge w$? The answer is no, just as not every tensor in $V \otimes V$ is of the form $v \otimes w$ (when $\dim V \geq 2$).
Example: let $V$ be a $4$-dimensional vector space with basis $e_1, e_2, e_3, e_4$. In $\Lambda^2(V)$, which is 4-dimensional with basis $e_i \wedge e_j$ for $i < j$, the sum $\omega := e_1 \wedge e_2 + e_3 \wedge e_4$ does not have the form $v \wedge w$ for some $v$ and $w$ in $V$.
To prove that, we describe a property that all $v \wedge w$ have but $\omega$ does not have: under the bilinear mapping $\Lambda^2(V) \times \Lambda^2(V) \to \Lambda^4(V)$ where $(v \wedge w, v' \wedge w') \mapsto v \wedge w \wedge v' \wedge w'$ on pairs of elementary wedge products in $\Lambda^2(V)$, we have $(v \wedge w, v\wedge w) \mapsto v \wedge w \wedge v \wedge w = 0$. However, $$ (\omega, \omega) \mapsto \omega \wedge \omega = (e_1 \wedge e_2 + e_3\wedge e_4) \wedge (e_1 \wedge e_2 + e_3\wedge e_4) = 2 e_1 \wedge e_2 \wedge e_3 \wedge e_4, $$ which is not $0$. Therefore $\omega$ is not of the form $v \wedge w$ for some $v$ and $w$ in $V$.