Centralizer of semisimple matrix Let $n>3$ be an integer. I have trouble finding a semisimple matrix $A\in M_n(\mathbb R)$ (i.e diagonalizable in $M_n(\mathbb C)$) such that its centralizer $Cent(A)=\{X\in M_n(\mathbb R)\;:\;AX=XA\}$ is isomorphic to $\mathbb C$. 
Or more general. If $\mathbb K$ is a field extension of $\mathbb F$ of degree $n$, what semisimple matrices in $M_n(\mathbb F)$ have centralizer isomorphic to $\mathbb K$?
 A: There is no such matrix. 
First, note that for any Lie algebra ${\mathfrak g}$ over a field ${\mathbb K}$, any $x\in{\mathfrak g}$ and any field extension ${\mathbb F}/{\mathbb K}$ we have $${\mathfrak z}_{\mathfrak g}(x)\otimes_{\mathbb K}{\mathbb F}\xrightarrow{\cong} {\mathfrak z}_{{\mathfrak g}\otimes_{\mathbb K}{\mathbb F}}(x),$$ so that for studying the dimensions of the centralizer ${\mathfrak z}_{\mathfrak g}(x)$ of $x$ in ${\mathfrak g}$ we may pass to a field extension.
In particular, to study the dimensions of centralizers of semisimple elements $x\in{\mathfrak g}{\mathfrak l}_n({\mathbb R})$ we may pass to ${\mathbb C}$ and consider the dimensions of centralizers of diagonalizable matrices $x\in {\mathfrak g}{\mathfrak l}_n({\mathbb C})$ instead. Replacing $x$ by a conjugate does not change the dimension of the centralizer, so we may assume that $x$ is diagonal, say $x = \text{diag}(x_1,...,x_n)$. Then you can check that $$\dim_{\mathbb C}{\mathfrak z}_{{\mathfrak g}{\mathfrak l}_n({\mathbb C})}(x)\ =\ n + \#\{i\neq j\ |\ x_i = x_j\}\geq n.$$
In particular, for $n>2$ no semisimple matrix over ${\mathbb R}$ can have centralizer ${\mathbb R}$-isomorphic to ${\mathbb C}$ for dimension reasons. For $n=2$, you can take $x := \tiny\begin{pmatrix} 0 & -1 \\ 1 & 0\end{pmatrix}$ and get the $2$-dimensional centralizer ${\mathbb R}\{1,x\}$ which as an ${\mathbb R}$-algebra is isomorphic to ${\mathbb C}$.
