Filling the gap of the lemma about bilinear map $\text{Lin}^2(\mathbf{R}^n, \mathbf{R})$ and endomorphism space End$(\mathbf{R}^n)$

Question

In Kolk's Multidimensional Real Analysis I: Differentiation

He proved the following lemma 2.9.1 to help find the Hessian matrix:

The proof he gave is simple:

The following are Lemma 2.7.4 and 2.6.1:

However, I found there are big gaps in the proof.

For $T \in \text{Lin}^2(\mathbf{R}^n,\mathbf{R})$, from Lemma 2.7.4, there is a linear isomorphism $v$, such that $T(h,k) = (v(T)h)k, v(T)\in \text{Lin}(\mathbf{R}^n, \text{Lin}(\mathbf{R}^n, \mathbf{R}))$.

Then from Lemma 2.6.1, we can find another linear isomorphism $\mu$, such that $T(h,k) = (v(T)h)k = \langle \mu(v(T)h), k \rangle$. Note that $v(T)h \in \text{Lin}(\mathbf{R}^n, \mathbf{R})$, so we can apply lemma 2.6.1. How does $\langle \mu(v(T)h), k \rangle$ become $\langle \lambda(T)h, k \rangle$ where $\lambda$ is again a linear isomorphism?

Update 1: In fact, there is an intrinsic relationship between a general bilinear form and a linear operator, see this question: But I still don't know how to get some insight from it.

Answer 1

As @Thorgott pointed out in the comment. I am almost there. We have: $ v(T)\in \text{Lin}(\mathbf{R}^n, \text{Lin}(\mathbf{R}^n, \mathbf{R}))$. We also have $\mu \in \text{Lin}(\text{Lin}(\mathbf{R}^n, \mathbf{R}),\mathbf{R}^n)$. So by defintion of function composition , $\mu \circ v(T) \in \text{Lin}(\mathbf{R}^n,\mathbf{R}^n) = \text{End}(\mathbf{R}^n)$. Here, $\mu \circ v(T)$ is $ \lambda(T)$. $\mu$ and $v$ is linear isomorphism, so is $\lambda$.