# On the definition of adjoint maps on inner product spaces

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.

• What's $\bar A$ over an arbitrary field $\mathbb F$? Dec 6 '20 at 12:47
• Fair point, I don't know. I think we are specializing to $\mathbb{F} = \mathbb{C}$ here. Dec 6 '20 at 13:18
• When $\mathbb F=\Bbb C$ you have $V=\Bbb C^n$, right? Dec 6 '20 at 13:28

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 $$x\mapsto A\cdot x$$.
Let $$e_1,\dots,e_n$$ be the standard basis, then $$\langle e_i,Ae_j\rangle=A_{i,j}$$ is the $$i,j$$th matrix entry, so $$(A^*)_{i,j}\ =\ \langle e_i,A^*e_j\rangle\ =\ \overline{\langle A^*e_j,e_i\rangle}\ =\ \overline{\langle e_j,Ae_i\rangle}\ =\ \overline{A_{j,i}}\,.$$
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 $$A^T$$.