Matrix whose characteristic polynomial does not splits over field. I need an example of matrix which is non-zero and not nilpotent whose characteristic polynomial does not splits over field $\mathbb{R}$

$$
\left(
\begin{array}{rr}
0 & -1 \\
1 & 0
\end{array}
\right)
$$
Thursday morning: if we try to diagonalize over the reals, we have real numbers $a,b,c,d$ with $ad-bc \neq 0,$ with
$$
\frac{1}{ad-bc}  \; \; 
\left(
\begin{array}{rr}
d & -b \\
-c & a
\end{array}
\right)
\left(
\begin{array}{rr}
0 & -1 \\
1 & 0
\end{array}
\right)
\left(
\begin{array}{rr}
a & b \\
c & d
\end{array}
\right) =
\frac{1}{ad-bc} \; \;
\left(
\begin{array}{rr}
-(ab+cd) & -(b^2 + d^2) \\
a^2 + c^2 & ab+cd
\end{array}
\right)
$$
The resulting matrix cannot be diagonal unless both $b^2 + d^2 = 0$ and $a^2 + c^2 = 0,$ meaning all four of $a,b,c,d = 0,$ with the contradictory outcome that $ad-bc = 0.$
With the same letters, if we now allow complex values, we can simply assign $a=d=1$ and $b=c=i$ to diagonalize over the complexes.
