Full-rank matrix I have a sparse square matrix and want to see if it is full rank (so that I can apply the implicit function theorem).
$$\left[\begin{array}{cccccccccc}
 0 & 1 & 1 & 1 & 0 & 0 & 0 & 0 & 0 & 0\\
 0 & 0 & 0 & 0 & 1 & 1 & 1 & 0 & 0 & 0\\
 x_{1}^{2} & Nx_{1} & 0 & 0 & -1 & 0 & 0 & 0 & 0 & 0\\
 0 & c & 0 & 0 & 0 & 0 & 0 & -x^2_{1} & 0 & 0\\
 x_{2}^{2} & 0 & Nx_{2} & 0 & 0 & -1 & 0 & 0 & 0 & 0\\
 0 & 0 & c & 0 & 0 & 0 & 0 & 0 & -x^2_{2} & 0\\
  0 & 0 & 0 & 0 & z_1 & 0 & 0 & -1 & 1 & 0\\
  x_{3}^{2} & 0 & 0 & Nx_{3} & 0 & 0 & -1 & 0 & 0 & 0\\
  0 & 0 & 0 & c & 0 & 0 & 0 & 0 & 0 & -x^2_{3}\\
  0 & 0 & 0 & 0 & z_2 & z_2 & 0 & 0 & -1 & 1
\end{array}\right]$$
where all variables are strictly positive, and $\sum x_i=1$.
Given that it is sparse, one approach that we considered is to do row-reductions and rearranging to reduce it to a block matrix. This is possible and yields:
$$\left[\begin{array}{ccc}
A & B & 0 \\
0 & C & D \\
E & 0 & F\end{array}\right]=\left[\begin{array}{ccc|ccc|ccc}
 N & 0 & 0 & x_{1}-1 & x_{1} & x_{1} & 0 & 0 & 0\\
 0 & N & 0 & x_{2} &x_2-1 & x_{2} & 0 & 0 & 0\\
 0 & 0 & N & x_{3} & x_{3} & x_3-1 & 0 & 0 & 0\\ \hline
 0 & 0 & 0 & x_1z_1 & 0 & 0 & -1 & 1 & 0\\
 0 & 0 & 0 & x_1z_2 & x_2z_2 & 0 & 0 & -1 & 1\\
 0 & 0 & 0 & x_1 & x_2 & x_3 & 0 & 0 & 0\\ \hline
 c & 0 & 0 & 0 & 0 & 0 & -x_{1}^2 & 0 & 0\\
 0 & c & 0 & 0 & 0 & 0 & 0 & -x_{2}^2 & 0\\
 0 & 0 & c & 0 & 0 & 0 & 0 & 0 & -x_{3}^2
\end{array}\right]$$

Thanks very much to Fred Hucht for getting me to think about $c$ and its relationship with $N$.
The following is an approach for small $c$:
As suggested by the mention of the implicit function theorem, the variables $x_1,x_2,x_3,z_1,z_2,N$ are implicit solutions to a complicated set of equations and $c$ is a parameter.
Above it was mentioned that the variables are strictly positive, but furthermore, they do not converge to $0$ as $c \rightarrow 0$ nor do they blow up.
This means that for $c$ sufficiently small, the above matrix is "almost" block diagonal and we can use the Gershgorin circle theorem to bound the eigenvalues away from $0$ (or we could argue continuity as $c\rightarrow 0$, or weight rows/columns in such a way that the matrix is diagonally dominant).
So, this resolves the "small" $c$ case. We think that the result is generically true for any $c$, but that remains open.
