how is this a cone?
$$a^2+bc=0$$ $$ab+bd=0$$ $$ac+cd=0$$ $$cb+d^2=0$$
I obtained this from $$A^2=\begin{bmatrix}a^2+bc&ab+bd\\ac+cd&cb+d^2\end{bmatrix}=0$$ and I know that the trace and det of $A$ are $0$, and apparently this is a cone said a friend. How is that a cone??
$A=\begin{bmatrix} a&b\\c&-a\end{bmatrix}$ and $A^2 = \begin{bmatrix}0&0\\0&0\end{bmatrix}=\begin{bmatrix}a^2-a^2&ab-ba\\ac-ca&-d^2+d^2\end{bmatrix}$