I'm reading the book "Int. to Applied Nonlinear Dynamical Systems and Chaos By Stephen Wiggins" and this book find the following affirmation.
We also show in this appendix that matrices with Jordan canonical form given by
$$\left(\begin{matrix}0&1\\0&0 \end{matrix}\right)$$ form a two-dimensional surface.
But I don't know how prove this affirmation. Some hint? I think in the action $ Ad: GLn(\mathbb{R})^{n}\times M_{2\times 2}(\mathbb{R})\to M_{2\times 2}(\mathbb{R}) $ given by $$Ad(m)= gmg^{-1},$$ where $GLn(\mathbb{R})$ is all nonsingular $n×n$ matrices, and to see that the orbit a matrix $m$ is a manifold, but I don't know.
Thanks in advance.
Such matrices
$$A=\begin{pmatrix}a&c\\b&d\end{pmatrix}$$
have $0$ as a double eigenvalue. Therefore :
$$\operatorname{tr}(A)=\lambda_1+\lambda_2=0=a+d \tag{1}$$
and
$$\det(A)=\lambda_1\lambda_2=0=ad-bc \tag{2}$$
constituting the intersection of 3D "hypersurface" given by (2) by an hyperplane given by (1).
As a consequence, we have a surface with dimension $2$.
Said otherwise, if we plug $d=-a$ into (2) we have: $a^2=-bc$ (think to $z^2=-xy$ in our ordinary space which is the equation of a surface)