Picture of set of nilpotent $2 \times 2$ matrices over $\mathbb{R}$ What does the set of nilpotent $2 \times 2$ matrices over $\mathbb{R}$ look like?
 A: Ultimately, what you're asking for is a picture of a $2$-dimensional subset of a $4$-dimensional space, so no picture is going to help you much here.  However, here are two helpful way of looking at this:
View 1: Every non-zero nilpotent $2 \times 2$ matrix is similar to the matrix $N = \pmatrix{0&1\\0&0}$.  In particular, if $A \neq 0$ is nilpotent and $2 \times 2$, then there is an invertible matrix $S$ such that $A = SNS^{-1}$.
So, one answer to your question is that if you apply the functions $f_S(X) = SXS^{-1}$ to $N$ for every $S$, then you get the set of (non-zero) nilpotent matrices.
View 2: Every $2 \times 2$ nilpotent matrix can be written in the form
$$
A = uv^T
$$
where $u$ and $v$ are two mutually orthogonal vectors.
Some other notable facts:


*

*The set of nilpotent matrices is closed under scalar multiplication.  In particular, if $A$ is nilpotent, so is $rA$ for any $r \in \Bbb R$.  So, the set consists of infinitely many rays emanating from the $0$ matrix.

*The set of nilpotent matrices is unbounded.

*The set of nilpotent matrices is precisely the set of matrices with trace and determinant $0$.

*The set of nilpotent matrices is closed. That is, all matrices on the "border" of this set are also nilpotent.

*The set of nilpotent matrices is (almost) a $2$-dimensional manifold in $4$-dimensional space (except for its behavior at $0$).

A: If $A\in\mathcal M_2(\Bbb R)$ a nilpotent matrix then $A=0$  or $A$ has index of nilpotence $2$ i.e. $A\ne0$ and $A^2=0$ so $\{0\}\ne Im A\subset \ker A$. Let $0\ne v\in Im A$ and then $v=Au$ for some $u\in\Bbb R^2$ and we prove easily that $(v,u)$ is a linearly independent family so it's a basis and the matrix $A$ on it take the form
$$\begin{pmatrix}0&1\\0&0\end{pmatrix}$$
