Differential Operators as vectors Do differential operators form a vector space and if so can we safely say that the del operator is a vector?
 A: Here is a new answer, which hopefully goes more into the direction the OP was asking: 
Asking whether something is a vector or not is not a meaningful question unless a vector space is specified. You can even go so far to say that everything is a vector and everything can be added, once you choose an appropriate vector space. For example both $\mathsf{Chicken}$ and $\mathsf{Cow}$ are elements of the free vector space $V$ generated by the set $\{\mathsf{Chicken},\mathsf{Cow}\}$ and $3\cdot \mathsf{Chicken} + 23\cdot \mathsf{Cow} $ is a well defined element in $V$ and thus a vector.
Once you have chosen a vector space $V$, you just call every element in $V$ vector. Of course you can still ask, whether differential operators lie in a vector space for which addition and scalar multiplication are actually useful. And to this the answer is yes: Below I describe some constructions of sensible vector spaces which contain differential operators.

First you need to say what a differential operator is to you, for the sake of simplicity I will focus on linear DOs with smooth coefficients.
Let $U\subset\mathbb{R}^n$ open and $p\in\mathbb{C}[x_1,...,x_n]$ a polynomial, then $p(D)$ is the differential operator  given by replacing the monomials $x_i^k$ of $p$ by $\partial_i^k$. For example if $p(x_1,x_2)=-x_1^2-x_2^2$, then $p(D)$ is the Laplace-operator on $\mathbb{R}^2$. Then
$$
V=\{p(D)\vert p \in\mathbb{C}[x_1,\dots,x_n]\}
$$
is obviously a vector space (with the same addition and scalar multiplication as for polynomials), it even is a complex algebra. If you want differential operators with nonconstant coefficients you can take polynomials $p\in C^{\infty}(U)[x_1,\dots,x_n]$ (here $C^\infty(U)$ is viewed as a ring).
In both cases there is an linear embedding $F\colon V\hookrightarrow \mathscr{L}(C_c^\infty(U),C_c^\infty(U))$ ($\mathscr{L}$ denotes the vector space of continuous and linear maps) where $F(p(D))u=p(D)u$ (the obvious application of the operator). The viewpoint that differential operators are a subspace of $\mathscr{L}(C_c^\infty,C_c^\infty)$ can be adapted in more general contexts: If $M$ is a smooth manifold then you can define the subspaces $\mathrm{Diff}^k(M)\subset \mathscr{L}(C_c^\infty(M),C_c^\infty(M))$ of linear DOs of order $k$ inductively by $\mathrm{Diff}^{-1}(M)=0$ and
$$
\mathrm{Diff}^{k+1}(M)=\{P\in \mathscr{L}(C_c^\infty(M),C_c^\infty(M))\vert \forall u\in  C^\infty(M): [P,u]\in \mathrm{Diff}^k(M)\},
$$
where $[P,Q]=PQ-QP$ is the commutator of operators and $u$ is viewed as operator that multiplies functions with $u$.
