Nilpotent matrices and idempotents are closed

Idea. Sets defined by polynomial equations are closed (preimages of $\{0\}$ under polynomial maps). Emptiness of the interior needs an actual perturbation argument.

Step 1: nilpotents are closed. $A$ is nilpotent iff $A^n=0$ (a nilpotent $n\times n$ matrix satisfies $A^n=0$ by Cayley-Hamilton, since its characteristic polynomial is $t^n$). The map $A\mapsto A^n$ is polynomial in the entries (hence continuous), so
$$\mathcal N=\{A:A^n=0\}=(A\mapsto A^n)^{-1}(\{0\})$$
is closed.

Step 2: nilpotents have empty interior. Let $N\in\mathcal N$ and $\varepsilon>0$. The eigenvalues of $N$ are all $0$, so those of $N+\varepsilon I$ are all $\varepsilon\neq0$: the matrix $N+\varepsilon I$ is invertible. A nilpotent matrix is singular, so $N+\varepsilon I\notin\mathcal N$, though $\|N+\varepsilon I-N\|=\varepsilon$. Hence no ball around $N$ lies in $\mathcal N$.

Step 3: idempotents are closed. $\mathcal P=\{P:P^2-P=0\}$ is the preimage of $\{0\}$ under the continuous map $P\mapsto P^2-P$.

Example. For $N=\begin{pmatrix}0&1\\0&0\end{pmatrix}$: $N^2=0$, while $(N+\varepsilon I)^2=\begin{pmatrix}\varepsilon^2&2\varepsilon\\0&\varepsilon^2\end{pmatrix}\ne0$.

Remark.
- Idempotents also have empty interior: if $P^2=P$ then all eigenvalues of $P$ lie in $\{0,1\}$, whereas $P+\varepsilon I$ has eigenvalues in $\{\varepsilon,1+\varepsilon\}$, so $P+\varepsilon I$ is not idempotent for $\varepsilon\ne0$ small, though it is at distance $\varepsilon$ from $P$. They are also *unbounded* for $n\ge2$: $\begin{pmatrix}1&t\\0&0\end{pmatrix}^2=\begin{pmatrix}1&t\\0&0\end{pmatrix}$ for all $t$ (oblique projections).
- The nilpotent cone is the fibre $\chi^{-1}(t^n)$ of the characteristic polynomial map (see Part 3).