In $\mathbb C^2$, assume that $<x,y>=x^tJ\bar y$ is an inner product. For $A \in M_{2\times 2} (\mathbb C)$, let $(L_A)^*=L_B$ where $L_A$ is the linear transformation of $\mathbb C^2$ such that $L_A \begin{pmatrix} z_1 \\ z_2 \\ \end{pmatrix} =A \begin{pmatrix} z_1 \\ z_2 \\ \end{pmatrix}$ and $(L_A)^*$ is its adjoint with respect to the given scalar product.
(a) When $J$ is the identity matrix, show that B=A*.
(b) When $J= \begin{pmatrix} 1 & i \\ -i & 2 \\ \end{pmatrix} $, find the explicit form of B.
What I want to know:
(a) Is it $B=A^*=I$? Do I have to choose a basis to show this?
(b) I don't have the faintest idea how to start. Would you give me a basic guideline to find the form of B?
Thanks in advance!
(Assuming that your notation is as I guessed in my comment above,) Write $$ \left\langle L_Ax , y \right\rangle = x^t A^t J \overline y, $$ and, denoting with $\overline B$ the entrywise complex conjugation, $$\left\langle x, L_B y \right\rangle = x^t J\overline B \overline y.$$ For $L_B=L_A^\star$ you need that the right hand sides of those expressions be equal. Since $x, y\in\mathbb C^2$ are arbitrary, this can only happen if...
Proceeding this way you can find an explicit expression for the matrix $B$ associated to $L_A^\star$. This expression should reduce to something familiar when $J$ is the identity matrix.
(These things arise in Lorentzian geometry, where one usually has $J=\mathrm{diag}(+1, -1\ldots -1)$ or $J=\mathrm{diag}(-1, +1\ldots +1)$).