Let $V$ be an $n$ dimensional vector space with a symmetric bilinear form induced by the matrix $A$, that is, $a_{ij} = (v_i, v_j)$ where $v_i$ are a basis of $V$.
I am stuck trying to show that $\text{rank}(A) = \text{dim}(V) - \text{dim}(\text{radical}(V))$. The thing that I am hung up on is the fact that the matrix is not interpreted as a linear map. If $A$ was just a linear map so that it operated on $V$ like $v \mapsto Av$ then it would follow from rank nullity theorem that $\text{rank}(A) = \text{dim}(V) - \text{dim}(\text{ker}(A))$.
But as it is I face two obstacles: The matrix acts on $V \times V$ via $(u,v)\mapsto u^T Av$ and the radical is defined as vectors $x$ such that $x^T Av = 0$ for all $v \in V$ which makes it difficult to see what the dimension of the radical might be.
How can I go about proving this?
Define the map $\Phi: V \to V^*$ (where $V^*$ is the dual space) given by $\Phi(x)= \phi_x$ where $\phi_x(v) = x^T A v$ (as in the comment above).
By definition, the radical of $A$ is $\ker \Phi$. Applying the rank-nullity theorem to $\Phi$ provides the equality that you want to prove.