What is the space $\operatorname{Sym}^2(V)$ and how does it act on the vector space $V$? 
If $V$ is a vector space over $\mathbb{C}$ with basis vectors $e_i$, what is the space $\operatorname{Sym}^2(V)$?

I am hoping someone can give me some insight into this space; perhaps by describing what it does to basis vectors?
Also, I have been trying to research more about it, but am unsure of its name, is it the symmetric space?
Thank you , this would be great for me to get some clarity on
 A: Since there are already two good answers I'll just add one thing that I think is important intuition:
If $V =W^*$ then by definition $V$ is the space of linear functions on $W$, in other words degree $1$ homogeneous polynomial functions on $W$. $Sym^2(V)$ is then just the space of degree $2$ homogeneous polynomial functions on $W$.
More explicitly if $W$ has a basis $e_1,...e_n$ then $V$ gets the dual basis $x_1,...,x_n$ which are just the coordinate functions on $W$. $Sym^2(V)$ is then just degree two homogenous polynomials in the variables $x_i$.
A: $Sym^2(V)$ has a basis consisting of elements
$$
e_i \vee e_j
$$
where $\vee$ denotes a symmetric tensor product.  In particular, $v \vee w = w \vee v$, and $v \vee (ax + by) = a v \vee x + b v \vee y$ for vectors $v,w,x,y$ and scalars $a,b$.
In my usage, I've heard $Sym^2(V)$ referred to as a "symmetric tensor product".

If $V$ has dimension $n$, the dimension of $Sym^k(V)$ is $\binom{n+k-1}{k}$.
A: The space $\operatorname{Sym}^2 \!V$ (sometimes also denoted $S^2 V$ or $\odot^2 V$) consists of all of the symmetric, contravariant rank-$2$ tensors on $V$.
A given basis $({\bf e}_a)$ of $V$ determines a basis of $\operatorname{Sym}^2 \!V$, namely,
$$({\bf e}_a \odot {\bf e}_b),$$
where $1 \leq a \leq b \leq b$. Here, $\odot$ is the symmetric tensor product, defined by
$${\bf x} \odot {\bf y} = \tfrac{1}{2}({\bf x} \otimes {\bf y} + {\bf y} \otimes {\bf x}) .$$
In particular, if $V$ is finite dimensional (say, $n := \dim V$), this shows that $\dim \operatorname{Sym}^2 \!V = \tfrac{1}{2} (n + 1) n$.
There is no natural action of $\operatorname{Sym}^2 \!V$ on $V$. If $V$ is equipped with a distinguished nondegenerate bilinear form $g$, however, that form determines an injective map $\flat : \operatorname{Sym}^2 \!V \to \operatorname{End} V$ and hence enables us to define an action of $\operatorname{Sym}^2 \!V$ on $V$, namely, $A \cdot {\bf v} := A^{\flat}({\bf v})$.
Note that all of this works just as well for any vector space over any field (except that over fields of characteristic $2$, we cannot divide by $2$ and so must adjust our definition of $\odot$ accordingly).
