Any connection between a symmetric bilinear form and a linear operator?

I will outline how to go from a symmetric bilinear form to a linear operator.

Given a symmetric bilinear form $\sigma : V \times V \to F$, consider the map $L' : V \to V^*$ given by $L'(v) = \sigma(v, \cdot)$. It is not a linear operator as the codomain is $V^*$ not $V$. However, as $V$ is finite dimensional, $V^*$ is isomorphic to $V$; let $\phi : V^* \to V$ be an isomorphism. Now it is not hard to check that $L = \phi\circ L'$ defines a linear operator on $V$.

It is worth noting that the isomorphism between $V$ and $V^*$ is not canonical, you need to choose a basis for $V$ in order to define one.

Every bi-linear form $\sigma:V\times V \to F$ defines a linear operator $A$ from $V$ to $V^*$; $A$ maps vector $v\in V$ to linear functional $l_v$ such that $l_v(u) = \sigma(u,v)$. The correspondence between $\sigma$ and $A$ is “canonical” — it doesn't depend on the basis.

Now suppose that we fix a basis $e_1, \dots, e_n$ in $V$. Define a basis $e_1^*, \dots, e_k^*$ in $V^*$ as follows: $e_i^*(e_i) = 1$ and $e_i^*(e_j) = 0$ if $j\neq i$. Let $\varphi$ be the map from $V^*$ to $V$ that sends $e_i^*$ to $e_i$. Then the operator $[L]_{B}$ is equal to $\varphi A$. Note that unlike the definition of $A$, the definition of $\varphi$ depends on the choice of the basis in $V$.