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.

  • 1
    $\begingroup$ What's $\bar A$ over an arbitrary field $\mathbb F$? $\endgroup$
    – cqfd
    Dec 6, 2020 at 12:47
  • $\begingroup$ Fair point, I don't know. I think we are specializing to $\mathbb{F} = \mathbb{C}$ here. $\endgroup$ Dec 6, 2020 at 13:18
  • $\begingroup$ When $\mathbb F=\Bbb C$ you have $V=\Bbb C^n$, right? $\endgroup$
    – cqfd
    Dec 6, 2020 at 13:28

2 Answers 2


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 $A^T$.

  • $\begingroup$ 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. $\endgroup$
    – Jhon Doe
    Apr 7, 2022 at 7:50
  • $\begingroup$ Yes, it applies to any subfield of $\Bbb C$. About completeness over such subfield, I'm not entirely sure. $\endgroup$
    – Berci
    Apr 7, 2022 at 9:28
  • $\begingroup$ Thanks. But just to clarify, the adjoint or transpose maps can still be defined right? $\endgroup$
    – Jhon Doe
    Apr 7, 2022 at 9:44
  • $\begingroup$ Yes. ${{{{})}}$ $\endgroup$
    – Berci
    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.

  • $\begingroup$ (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.) $\endgroup$ Feb 26, 2023 at 19:55
  • $\begingroup$ (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.) $\endgroup$ Feb 26, 2023 at 20:05

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