I'd like to know if there is a 4x4 matrix with real entries that can be transformed to the following matrix A. $x,y\in\mathbb{R}, i=\sqrt{-1}$,$$ \ $$ $A=\begin{bmatrix} (x+iy) & 1 & 0 & 0\\0 & (x+iy) & 0 & 0\\0 & 0 & (x-iy) & 0\\0 & 0 & 0 & (x-iy)\end{bmatrix}$.
Specifically, does there exist a $B\in M_{4,4}(\mathbb{R})$ such that $Q^{-1}.B.Q=A$ where $Q$ is a transformation matrix, $Q\in M_{4,4}(\mathbb{C})$.
Any suggestions to the problem are welcome! ^^
P.S. This question came to me when I was trying to list all the possible Segre Types for a 4x4 matrix.
Note that if $A$ is similar to a real matrix $B$, that is, if $B=PAP^{-1}$ for some invertible matrix $P$, then $PAP^{-1} =B=\overline{B}=\overline{PAP^{-1}}=(\overline{P})\overline{A}(\overline{P})^{-1}$. Thus $A$ must be similar to $\overline{A}$.
However, when $y\ne0$, your $A$ is not similar to $\overline{A}$ because the geometric multiplicity of $x+iy$ is $1$ in $A$ but $2$ in $\overline{A}$. Therefore $A$ is not similar to a real matrix in this case.