# 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 $\mathbb F$?
– cqfd
Dec 6, 2020 at 12:47
• Fair point, I don't know. I think we are specializing to $\mathbb{F} = \mathbb{C}$ here. Dec 6, 2020 at 13:18
• When $\mathbb F=\Bbb C$ you have $V=\Bbb C^n$, right?
– cqfd
Dec 6, 2020 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 $$A^T$$.

• Hello, can this natural inner product can be used for any subfield of $\mathbb{C}$? I suppose its possible but the resulting inner-product space need not be complete i.e. Hilbert. Am I correct in assuming that we can still define the adjoint or transpose on such spaces as no notion of completeness is needed? Thanks. Apr 7, 2022 at 7:50
• Yes, it applies to any subfield of $\Bbb C$. About completeness over such subfield, I'm not entirely sure. Apr 7, 2022 at 9:28
• Thanks. But just to clarify, the adjoint or transpose maps can still be defined right? Apr 7, 2022 at 9:44
• Yes. ${{{{})}}$ Apr 7, 2022 at 10:14

If $$V$$ is an "abstract" finite-dimensional vector space over $$\mathbb{C}$$, the matrix $$A$$ corresponding to a linear transformation self-map $$L: V \to V$$ depends on the ordered basis one uses for $$V$$. In particular, if $$B_1 = (v_1, \ldots v_n)$$ is one ordered basis for $$V$$ and $$A$$ is the matrix for $$L$$ in the ordered basis $$B_1$$, $$B_2 = (w_1, \ldots w_n)$$ is another ordered basis for $$V$$ and $$B$$ is the matrix for $$L$$ in the ordered basis $$B_2$$, and $$P$$ is the transition matrix with $$AP = PB$$, then $$B^* = (P^{-1}AP)^* = P^*A^*(P^{-1})^*$$, which is not $$P^{-1}A^*P$$, the matrix for $$B^*$$ with respect to the ordered basis $$B_2^*$$, unless $$P$$ is unitary. (In fact, the ordered basis $$B_2^*$$ isn't exactly well-defined at this point; see below.)

This is to say, if one has a linear transformation $$L$$, one chooses an orthonormal ordered basis $$U_1 = (u_1, \ldots, u_n)$$ for $$V$$, sets $$R$$ to be the matrix for $$L$$ in the ordered basis $$U_1$$, chooses $$U_2 = (x_1, \ldots, x_n)$$ to be another orthonormal ordered basis for $$V$$, and sets $$S$$ to be the matrix for $$L$$ in the orthonormal ordered basis $$U_2$$, so that $$RQ = QS$$ for a unitary transition matrix $$Q$$, then $$S^* = Q^{-1}R^*Q$$ is the matrix for $$L^*$$ with respect to the basis $$U_2^* = (x_1^*, \ldots, x_n^*)$$.

So, one generally needs to have an inner product space for an "abstract" finite-dimensional vector space in order to have a well-defined notion for the matrix of the adjoint of a linear transformation, and, in this setting, the notion is only well-defined in the setting or category of orthonormal ordered bases for the inner product space being used to find the matrix of a linear transformation.

In general, (1) the notion of an eigenvalue and eigenvector is only well-defined for a self-map of a vector space to itself, and (2) the correspondence between the (conjugate) transpose of a matrix and the matrix corresponding the adjoint of the linear transformation defined by the matrix is only well-defined in the category of orthonormal ordered bases ("orthonormal frames") for the inner product spaces.

• (A frame, the way my linear algebra professor used the terminology, corresponding to an ordered basis $B = (v_1, \ldots, v_n)$ for a finite-dimensional vector space over $\mathbb{F}$ is the map $\Phi: \mathbb{F}^n \to V$ given by $\Phi(e_i) = v_i$; I don't know how standard that terminology is.) Feb 26, 2023 at 19:55
• (I guess the matrix $Q$ could be unitary without $U_1$ and $U_2$ being orthonormal, i,e., one could have $||u_i|| = ||x_i||$ and $\langle u_i\ |\ u_j\rangle = \langle x_i\ |\ x_j\rangle$ without either basis being orthonormal; I haven't completely thought about this case.) Feb 26, 2023 at 20:05