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.


As mentioned in the comments, it is required to assume $\mathbb{F} = \mathbb{C}$ for $\bar A$ to make sense.

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


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