Problem related to Complex and linear mappings Since I could not even solve part 1, I do not have any idea about part 2 ( I thought $K$ and $A$ are identical, but apparently they are not.)
Here is the question: 


*

*Let $V_1 =\mathbb{C^n}$ with the inner product $<x,y>  = y^*x$

*Let $V_2 = \mathbb{R^{2n}}$ with the inner product $<u,v> = v^Tu$

*Let $z$ be a mapping from $\mathbb{C^n}$ to $ \mathbb{R^{2n}}$, where: $z(x) =$ 
$$\begin{bmatrix}Re(x)\\Im(x)\end{bmatrix}$$

*Let $A \in \mathbb{C^{n*n}}$ define a linear mapping $f:\mathbb{C^n} \rightarrow \mathbb{C^n}$ s.t for any $x \in \mathbb{C^n}$, $f(x) = Ax$

*Let $g: \mathbb{R^{2n}} \rightarrow \mathbb{R^{2n}}$ is the mapping corresponding to $f$ s.t for any $x \in V_1$ $g(z(x)) = z(f(x))$
Now we can write linear mapping $g$ explicitly as: $g(u) = Ku$, $u \in \mathbb{R^{2n}}$
Question:
1) Write $K$ in terms of A.
2) If $K$ is Hermitian, what can we say about the relation between it and $A$'s eigenvalues and eigenvectors?
edit: 
I solved the first part:
$K$ =  $$\begin{bmatrix}A & 0\\0 & A\end{bmatrix}$$
any ideas about the second part?
 A: Note that $A = \operatorname{Re}A +i\operatorname{Im}A$. Also $$z(x) = \begin{bmatrix}\operatorname{Re}x\\\operatorname{Im}x\end{bmatrix}\qquad z^{-1}\left(\begin{bmatrix}v_1\\v_2\end{bmatrix}\right) = v_1 + iv_2$$
So, since $K = z\circ A\circ z^{-1}$,
$$\begin{align}K\begin{bmatrix}v_1\\v_2\end{bmatrix} &= zAz^{-1}\begin{bmatrix}v_1\\v_2\end{bmatrix}\\
&= z(A\operatorname{Re}A +i\operatorname{Im}A)(v_1 + iv_2)\\
&=z\left((\operatorname{Re}(A)v_1 - \operatorname{Im}(A)v_2) + i(\operatorname{Im}(A)v_1 + \operatorname{Re}(A)v_2)\right)\\
&=\begin{bmatrix}\operatorname{Re}(A)v_1 - \operatorname{Im}(A)v_2\\\operatorname{Im}(A)v_1 + \operatorname{Re}(A)v_2\end{bmatrix}\\
&=\begin{bmatrix}\operatorname{Re}A & - \operatorname{Im}A\\ \operatorname{Im}A & \operatorname{Re}A\end{bmatrix}\begin{bmatrix}v_1\\v_2\end{bmatrix}
\end{align}$$
Therefore,
$$K = \begin{bmatrix}\operatorname{Re}A & -\operatorname{Im}A\\ \operatorname{Im}A & \operatorname{Re}A\end{bmatrix}$$

Since $K$ is real, Hermitian = symmetric. So
$$\begin{bmatrix}(\operatorname{Re}A)^T & (\operatorname{Im}A)^T\\ -(\operatorname{Im}A)^T & (\operatorname{Re}A)^T\end{bmatrix} = K^T = K = \begin{bmatrix}\operatorname{Re}A & -\operatorname{Im}A\\ \operatorname{Im}A & \operatorname{Re}A\end{bmatrix}$$
So $\operatorname{Re}A$ is symmetric and $\operatorname{Im}A$ is skew-symmetric. Now $$A^* = (\operatorname{Re}A)^T -i(\operatorname{Im}A)^T = \operatorname{Re}A +i\operatorname{Im}A = A$$
Therefore $K$ is Hermitian $\implies$ $A$ is Hermitian, and therefore the eigenvalues of $A$ are real. Now if $\lambda$ is real, then $z(\lambda x) = \lambda z(x)$. So
$$Ax = \lambda x \iff z(Ax) = z(\lambda x)\iff Kz(x) = \lambda z(x)$$
So the eigenvalues of $A$ and $K$ are the same, and their eigenspaces are the images of each other under the transformations $z$ and $z^{-1}$.
