Matrix sends a vector to a scalar multiple of its complex conjugate Does anyone know any methods for solving the following for $\mathbf{v}$:
$\mathbf{A\bar{v} = \lambda v}$
Ideas:

*

*Maybe multiplying by some matrix takes the complex conjugate

*This is like a weird eigenvector problem where instead of sending the vector to a scalar multiple of itself, it sends it to a scalar multiple of its complex conjugate

*What if we force $\mathbf{v} = e^{i\mathbf{k}}$ for some $\mathbf{k}$? Does this help?

 A: Consider everything to be over a $2n$-dimensional real vector space, in the obvious way, ordering the entries as (real part of the first entry), (imaginary part of the first entry), (real part of the second entry), etc.
Then we can replace $\mathbf{A}$ with a new matrix $r(\mathbf{A})$ in the obvious way, $\mathbf{v}$ by $r(\mathbf{v})$ (in the same obvious way), $\bar{\mathbf{v}}$ by $\mathbf{B}r(\mathbf{v})$, where $$\mathbf{B} = \bigoplus^n \left(\begin{array}{cc}1 & 0 \\ 0 & -1\end{array}\right),$$ and $\lambda$ by a matrix $$r(\lambda) = \bigoplus^n\left(\begin{array}{cc}\mathfrak{Re}(\lambda) & \mathfrak{Im}(\lambda)\\-\mathfrak{Im}(\lambda) & \mathfrak{Re}(\lambda)\end{array}\right).$$
That then gives us a new equation of the form $(r(\mathbf{A})\mathbf{B})r(\mathbf{v}) = r(\lambda)r(\mathbf{v})$, which you can then solve in the usual way (getting eigenvalues/eigenvectors for $r(\mathbf{A})\mathbf{B}$). The eigenvectors so produced are the $r(\mathbf{v})$ for $\mathbf{v}$ solving the original equation, which you can recover by simply taking each $k$th pair $(x_k,y_k)$ of non-overlapping consecutive entries of each $r(\mathbf{v})$ and letting your $k$th entry of $v$ be $x_k + iy_k$.
A: $e^{i \mathbf k}$ doesn't make sense if $\mathbf k$ is a vector. $e^x$ is defined for rings, but unless there is some multiplicative operator defined on the vector space that $\mathbf k$ is in, that space is not a ring. We can define $e^{iA}$, since matrices are a ring, but I don't think that helps. Now, if we take $\mathbf v = e^{ix}\mathbf u $ for some real-valued vector $\mathbf u$ and scalar $x$, we can say $\mathbf A e^{-ix}\mathbf u = e^{ix}\lambda \mathbf u$. Scalars commute with matrices, so we have $ e^{-ix}\mathbf A\mathbf u = e^{ix}\lambda \mathbf u$ and therefore $ \mathbf A\mathbf u = e^{2ix}\lambda \mathbf u$, which shows that $\mathbf u$ is an eigenvector of  $\mathbf A$ with eigenvalue $e^{2ix}\lambda $. So that implies that there is a real-valued eigenvector $ \mathbf u$, and we need to take its eigenvalue, write it as $e^{2ix}\lambda $, and then take $\mathbf v = e^{ix}\lambda \mathbf u $. But if an eigenvalue can be written as $e^{2ix}\lambda $,  then it also can be written as $e^{2(x+\pi)i}\lambda $, giving $\mathbf v = e^{(x+\pi)i}\lambda \mathbf u $.
