I'm trying to prove the uniqueness of the adjoint operator between two finite-dimensional Hilbert spaces $E$ and $F$. My idea is to simply use the isomorphism between finite-dimensional Hilbert spaces and the Euclidean spaces while also using the Euclidean inner product by its definition.
The origional problem:
Suppose $(E,(\cdot \mid \cdot )_{E})$ and $(F,(\cdot \mid \cdot )_{F})$ are two finite-dimensional Hilbert spaces. Verify that to each $A\in \mathcal{L}(E,F)$ there is a unique $A^*\in\mathcal{L}(F,E)$ , such that $\forall x\in E, y\in F$, $(Ax\mid y)_F = (x\mid A^*y)_E$.
Also there are following subquestion asking to prove the form of the $A^*$ is exactly $[\overline{a_j^k}]$, and so on.
My proof is like this:
Suppose $\dim E=n, \dim F=m$, and let $[A]=[a_k^j]\in\mathbb{K}^{m\times n}$ (the upper symbol is the line number), then $\forall x=\left( x^1, \cdots , x^n \right) \in E ,y=\left( y^1,\cdots ,y^m \right) \in F$,
\begin{align} \left( x\mid A^*y \right) _E&=\left( Ax\mid y \right) _F=\left( \left( \sum_{j=1}^n{a_{j}^{1}x^j}, \cdots ,\sum_{j=1}^n{a_{j}^{m}x^j} \right) \mid \left( y^1,\cdots ,y^m \right) \right) _F \\ &=\sum_{k=1}^m{\left( \sum_{j=1}^n{a_{j}^{k}x^j\overline{y^k}} \right)}=\sum_{j=1}^n{x^j\sum_{k=1}^m{a_{j}^{k}\overline{y^k}}} \\ &=\sum_{j=1}^n{x^j\overline{\sum_{k=1}^m{\overline{a_{j}^{k}}y^k}}} \\ &=\left( x\mid \left( \sum_{k=1}^m{\overline{a_{1}^{k}}y^k},\cdots ,\sum_{k=1}^m{\overline{a_{n}^{k}}y^k} \right) \right) \end{align}
, and therefore $\left[ A^* \right] =\left[ \overline{a_{j}^{k}} \right] \in \mathbb{K} ^{n\times m}$, which is unique.
However, I think it quite unrigorous to directly use the Euclidean inner product on two random Hilbert spaces. Can anyone examine this proof? Thanks!
Update:
Thanks to @Qiaochu Yuan's hint, now I finished the proof, writing the matrix in the standard form between the two orthonormal basis, while the rest are almost identical. Are there any other details to notice? And are there other approaches to proving it?
Suppose there exists $B^*$ such that it satisfies the adjoint property. Hence we have $\left<x,A^*y\right>_E =\left<x,B^*y\right>_E, \forall x,y \in E,F$. This means $\left<x,(A^*-B^*)y\right>_E=0$ $\forall x,y \in E,F$. In particular, for any fixed $y$, take $x = (A^*-B^*)y$. Then the inner product axioms would imply $A^*=B^*$.