Example of nilpotent matrix such that $0$ is not in the closure of the similarity class of $A$

Question

One can prove that if $A\in\mathcal{M}_n(\mathbb{C})$ is nilpotent then $0\in\overline{\mathcal{S}(A)}$. Where $\mathcal{S}(A)$ is the similarity class of $A$.

Note that $$\mathcal{S}_\mathbb{R}(A)=\mathcal{S}_{\mathbb{C}}(A)\cap \mathcal{M}_n(\mathbb{R}),$$ so $0\in\overline{\mathcal{S}_\mathbb{C}(A)}\cap\overline{\mathcal{M}_n(\mathbb{R})}$, unfortunately that doesn't mean that $0\in\overline{\mathcal{S}_\mathbb{C}(A)\cap\mathcal{M}_n(\mathbb{R})}$.

I don't think this holds in $\mathbb{R}$, but I'm struggling finding an exemple. I've tried with $n=2$ making explicite $\mathcal{S}(A)$ without any success.

Could you please help me ?

Answer 1

I was wrong, it holds over $\mathcal{M}_n(\mathbb{R})$, in the proof we only use that a nilpotent matrix is always triangulasible.