How to prove that the matrix of a symmetric bilinear form is symmetric A bilinear form $B:V \times V \rightarrow K$ is said to be symmetric if $B(\mathbf{v,w}) = B(\mathbf{w,v}) \ \forall \mathbf{v,w} \in V,$ where $V$ is a vector space over some field $K.$
A matrix $A$ is said to be symmetric $\iff$ $\langle Ax, y \rangle = \langle x, Ay \rangle \ \forall x, y \in V.$
When I try to prove this using those two facts, I come across issues such as doesn't $V$ need to be an inner product space to be able to define the notion of the standard inner product (dot product)? Also, (real) dot products map to $\mathbb{R}$ and not an arbitrary field $K.$ 
 A: For your second question, it's an important observation that the whole linear algebra formalism requires the numbers to have addition, subtraction, multiplication and division with any nonzero. 
It means that all works the same way when the scalars (coefficients, matrix entries, etc.) are taken from an 'over arbitrary field' $K$ in place of the reals $\Bbb R$.
This definition of a matrix being symmetric indeed presumes an inner product, but it can be reformulated simply in terms of the matrix entries: 
Apply the equation to the standard basis vectors $x=e_i,\,y=e_j$. Since the $\langle Ae_i,e_j\rangle$ is the $j$th coordinate of the $i$th column of $A$, and $\langle e_i,Ae_j\rangle$ is the $i$th coordinate of the $j$th column, it indeed means that the matrix is symmetric:
$$a_{ij}=a_{ji}$$
In the context of matrices, this definition (or its transpose form: $A^T=A$) doesn't require a priori any inner product, neither any vector space.
However, once we fix a basis $(e_i)$ in a finite dimensional vector space $V$, it automatically induces an inner product (by the usual formula with coordinates), in which $(e_i)$ becomes orthonormal. 
In your first paragraph, there is no basis and no inner product fixed. 
On the other hand, when we use matrices (with entries from $K$), we might automatically associate the vector spaces $K^n$ with their standard basis and inner product.
