I was finding the Jordan form for a matrix $A \in M_{4}(\mathbb{R})$ given by:
$$ A= \begin{pmatrix} 0 & -1 & -1 & -1 \\ 1 & 2 & 1 & 1 \\ 0 & 0 & 0 & -1 \\ 0 & 0 & 1 & 2 \end{pmatrix} \in M_{4}(\mathbb{R}) $$
whose characteristic polynomial is, $p_{A}(x)=(x-1)^{4}$ and minimal $m_{A}(x)=(x-1)^{2}$. And I had no problem in finding it's Jordan form:
$$ J= \begin{pmatrix} 1 & 0 & 0 & 0 \\ 1 & 1 & 0 & 0 \\ 0 & 0 & 1 & 0 \\ 0 & 0 & 1 & 1 \end{pmatrix} \text{ and, } P= \begin{pmatrix} 1 & -1 & 0 & -1 \\ 0 & 1 & 0 & 1 \\ 0 & 0 & 0 & -1 \\ 0 & 0 & 1 & 1 \end{pmatrix} \text{ such that } J=P^{-1}AP. $$
But that's not the point of my question, since the problem came in the next section as they asked me if $J \in M_{4}(\mathbb{R})$ would change in any way (diagonalize) if instead of working in $M_{4}(\mathbb{R})$, we worked in $M_{4}(\mathbb{C})$. I suspect that $J$ won't change since the characteristic polynomial factorizes in $\mathbb{R}_{4}[x]$ as $p_{A}(x)=(x-1)^{4}$, and as $\mathbb{R}\subset \mathbb{C}$ this implies that $p_{A}(x)$ will have the same factors in $\mathbb{C}_{4}[x]$ ($p_{A}(x)=(x-1)^{4}\in \mathbb{C}_{4}[x]$). So, I suspect that $J\in M_{4}(\mathbb{C})$ will also be
$$ J= \begin{pmatrix} 1 & 0 & 0 & 0 \\ 1 & 1 & 0 & 0 \\ 0 & 0 & 1 & 0 \\ 0 & 0 & 1 & 1 \end{pmatrix} \in M_{4}(\mathbb{C}) $$
but i'm not sure. If so, where is my flaw? Is it true in general?
Thanks in advanced.