# Understanding the association of the bilinear form and matrix associated with it

From S.L Linear Algebra:

Show that the association $$A \rightarrow g_A$$ is an isomorphism between the space of $$m \times n$$ matrices, and the space of bilinear maps of $$\mathbb{K}^m \times \mathbb{K}^n$$ into $$\mathbb{K}$$.

Note: In calculus, if $$f$$ is a function of $$n$$ variables, one associates with $$f$$ a matrix of second partial derivatives ($$\frac{\partial^2 f}{\partial x_i \partial x_j}$$), which is symmetric. This matrix represents the second derivative, which is a bilinear map.

$$g_A$$ simply implies that the bilinear map $$g$$ has a matrix $$A$$ associated with it, such that $$g_A(X, Y)=X^TAY$$.

# Confusion:

It is visible that $$A$$ is simply an arbitrary $$m \times n$$ matrix and $$g_A$$ is bilinear map $$\mathbb{K}^m \times \mathbb{K}^n \rightarrow \mathbb{K}$$

Association above $$A \rightarrow g_A$$, can be "expanded" to $$A \rightarrow \mathbb{K}^m \times \mathbb{K}^n \rightarrow \mathbb{K}$$, which makes it seem like a composite map $$L: A \circ \mathbb{K}$$.

Thus if we have $$X \in \mathbb{K}^m$$ and $$Y \in \mathbb{K}^m$$, then it sems like $$L(X, Y)=g_A(X^TAY)=(X^TAY)^2$$.

I'm highly confused about the definition of mapping $$L$$, if I'm certain that the definition that I've obtained is correct, then perhaps I could show that kernel of $$L$$ is trivial (showing injectivity), and then by rank nullity that map is bijective. Therefore, is the result above appropriate by any means?

Perhaps the example in the Note was a linear map similar to $$L$$ above (which is why second derivative is discussed there)?

It seems to me the Note is talking about specialization of hessian matrix which contains bilinear forms (second derivatives) as elements.

In short, what's the correct definition of association $$A \rightarrow g_{A}$$.

$$A\to g_A$$ takes an $$m\times n$$ matrix and maps it to the associated bilinear form. So it should be something of the form: $$L:\mathbb K^{m\times n}\to Bil_{\mathbb K}(\mathbb K^m\times \mathbb K^n) \\ A\longmapsto g_A$$ So $$L(A)(X,Y)=g_A(X,Y)$$.