1
$\begingroup$

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.

$\endgroup$
3
  • $\begingroup$ 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
2
$\begingroup$

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 $A^T$.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.