Definition: Let $A:V\to V$ be linear, where $V$ is an inner product space over $\mathbb{F}$. Then $A^*:V\to V$ is defined by the equation: $$\langle y,A^*x\rangle = \langle Ay,x\rangle \text{ for all }x,y\in V$$
Here $A^*$ is called the adjoint operator with respect to $A\in\mathcal{L}(V)$.
So, it is clear that $A^*$ is defined on inner product spaces only. However, we see that the conjugate transpose of a matrix is defined regardless of whether $V$ is an inner product space or not (right?), i.e. if $A\in M_n(\mathbb{F})$, we define $A^* = (\bar A)^T$.
I suspect a possible correlation between the conjugate transpose defined for matrices, and the adjoint operator defined on inner product spaces, because of results such as:
Let $\mathcal{B}$ be an orthonormal ordered basis of $V$, then $A:V\to V$ is self adjoint (i.e. $A^* = A$) iff $[A]^*_{\mathcal{B}} = [A]_{\mathcal{B}}$ (matrices of linear maps w.r.t. basis $\mathcal{B}$).
However, it seems weird that $A^*$ (adjoint) in the context of linear maps is defined only on inner product spaces, while the conjugate transpose of a matrix $A\in M_n(\mathbb{F})$ is defined more generally. Am I missing something here? I was hoping for a stronger connection between the two definitions.
Thanks!
P.S.
As mentioned in the comments, it is required to assume $\mathbb{F} = \mathbb{C}$ for $\bar A$ to make sense.
Best Answer
The point is that $\Bbb F^n$ carries a natural inner product, when $\Bbb F=\Bbb R$ or $\Bbb C$, namely the ordinary one: $$\langle x,y\rangle:=\sum_k\overline{x_k}y_k\,.$$ And one can prove that $A^*={\bar A}^T$ using this inner product, when identifying a matrix $A$ with the linear map $x\mapsto A\cdot x$.
Note that over $\Bbb R$, conjugation is the identity, we can omit it, so in that case the adjoint of $A$ is simply its transpose $A^T$.