With the definition $S^{n} = \{\ \mathbf{x} \in \mathbb{R}^{n+1}\ | \ ||\mathbf{x}|| = 1\ \}$, and the function $\ f:\mathbb{R}^{n} \to S^{n} \setminus \{ (0,...,0,1) \}$ defined by:
$f(\mathbf{u}) = \left( \ \frac{ 2 u_{1} }{ ||\mathbf{u}||^{2} + 1 },\ ...,\ \frac{ 2 u_{n} }{ ||\mathbf{u}||^{2} + 1 },\ \frac{ ||\mathbf{u}||^{2} - 1 }{ ||\mathbf{u}||^{2} + 1 } \ \right)$
I'm trying to show that the rank of the Jacobian matrix $\mathcal{J}$, is $n$ for all $\mathbf{u} \in \mathbb{R}^{n}$, and I can't figure out how to do this.
After some annoying computations, I find that the matrix looks like:
$\mathcal{J}(\mathbf{u}) = \frac{1}{\left( ||\mathbf{u}||^2 + 1 \right)^{2}} \left[ \begin{matrix} \ 2 \left( ||\mathbf{u}||^2 + 1 \right) - 4 u_{1}^2 & - 2 u_{1} u_{2} & \cdots & - 2 u_{1} u_{n} \\ - 2 u_{1} u_{2} & 2 \left( ||\mathbf{u}||^2 + 1 \right) - 4 u_{2}^2 & \cdots & - 2 u_{2} u_{n} \\ \vdots & \vdots & \ddots & \vdots \\ - 2 u_{1} u_{n} & - 2 u_{2} u_{n} & \cdots & 2 \left( ||\mathbf{u}||^2 + 1 \right) - 4 u_{n}^2\ \\ 4 u_{1} & 4 u_{2} & \cdots & 4 u_{n}\ \\ \end{matrix} \right]$
How do I proceed to show that $\mathrm{rank}(\mathcal{J}(\mathbf{u})) = n$ for all $\mathbf{u}$?
It's easy to see that $\mathcal{J}(\mathbf{0})$ has rank $n$. I have already proven that $f$ is a homeomorphism of class $C^{\infty}$, so I am wondering is there some result I can use that states that the rank remains $n$ in a neighbourhood of $\mathbf{0}$ in $\mathbb{R}^{n}$? Otherwise, how else can I prove this?
EDIT: I realize that I can row-reduce the matrix, but this seems tedious and I am looking for ways to avoid row-reducing
The stereographic projection $$s:S^{n}\setminus \{(0,...,0,1)\}\to \mathbb R^n: x=(x_1,\dots,x_{n+1})\mapsto u= \frac {1}{1-x_{n+1}}(x_1,\dots,x_n)$$ is clearly $C^\infty$ and is the inverse of your map: $s=f^{-1}$.
Hence your map $f:\mathbb{R}^{n} \to S^{n} \setminus \{ (0,...,0,1) \}$ is a diffeomerphism and as such has rank $n$.