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Well, I have the following inquiry: I have an element $g \in \mathrm{O}(V)$. I know that such element can be decomposed as the sum of two operators: $$g=p_g+q_g,$$ where $p_g$ is linear and $q_g$ is antilinear.

I’m are working over (finite) vector spaces over $\mathbb{C}$.

I’m trying to calculate $$T:=q_gp_g^{-1},$$ (assuming that $p_g$ is invertible) for a specific $g$. I already expressed $p_g$ in matrix form and calculated it’s inverse. However, since $q_g$ is not linear, it doesn’t hace a matrix representation. ¿How can I calculate $q_gp_g^{-1}$? Specifically, $$p\sim \begin{bmatrix} \alpha & -\overline\beta\\ \beta & \overline\alpha \end{bmatrix}$$ and $q=-\beta \overline{\beta}v+\overline\alpha \beta u$, where $\{ v,u\}$ is a basis of $V$ and $\alpha, \beta \in \mathbb{C}$. If $q$ were a linear transformation, I could put it in matrix form…

What’s the matrix form of an antilinear transformation?

Thanks in advance!

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    $\begingroup$ An antilinear map is still linear over $\mathbb{R}$. $\endgroup$
    – Qiaochu Yuan
    Commented Oct 20, 2022 at 2:10
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    $\begingroup$ You mention matrices, so I will assume $\Bbb C^n$. If $C:\Bbb C^n\to\Bbb C^n$ is complex conjugation, then any $\Bbb R$-linear map $A:\Bbb C^n\to\Bbb C^n$ is expressible as $A=A_0+A_1C$ where $A_0$ and $A_1$ are both $\Bbb C$-linear. (Being orthogonal is irrelevant.) Then $A_0(v)=\frac{1}{2i}(iA(v)+A(iv))$ and $A_1(v)=\frac{1}{2i}(iA(v)-A(iv))$. Not sure what form you have $A$ in that you can express $A_0,A_1$ with it. Not sure what conditions on $A_0,A_1$ are equivalent to $A^{-1}$ existing. $\endgroup$
    – anon
    Commented Oct 23, 2022 at 0:02
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    $\begingroup$ At the end of the question, there is something unclear. It seems that you explicitly provide p_g and q_g (but you forget the subscripts). While p is fine (almost... you provide a matrix that can easily be turned into a linear operator), q seems to be just a vector, since alfa and beta are (known?) parameters, and u and v are also known vectors, the basis. Instead, I expected an operator on V. The general question has been answered in the first comment (an antilinear map is linear if you represent the map over R). In order to answer about the specific case, I think that you should clarify. $\endgroup$
    – Doriano Brogioli
    Commented Oct 27, 2022 at 8:10

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