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:


$$\left[\begin{array}{ccc}
A & B & 0 \\
0 & C & D \\
E & 0 & F\end{array}\right]=\left[\begin{array}{ccc|ccc|ccc}
 N & 0 & 0 & x\_{1}-1 & x\_{1} & x\_{1} & 0 & 0 & 0\\
 0 & N & 0 & x\_{2} &x\_2-1 & x\_{2} & 0 & 0 & 0\\
 0 & 0 & N & x\_{3} & x\_{3} & x\_3-1 & 0 & 0 & 0\\ \hline
 0 & 0 & 0 & x\_1z\_1 & 0 & 0 & -1 & 1 & 0\\
 0 & 0 & 0 & x\_1z\_2 & x\_2z\_2 & 0 & 0 & -1 & 1\\
 0 & 0 & 0 & x\_1 & x\_2 & x\_3 & 0 & 0 & 0\\ \hline
 c & 0 & 0 & 0 & 0 & 0 & -x\_{1}^2 & 0 & 0\\
 0 & c & 0 & 0 & 0 & 0 & 0 & -x\_{2}^2 & 0\\
 0 & 0 & c & 0 & 0 & 0 & 0 & 0 & -x\_{3}^2
\end{array}\right]$$



Full-rank matrix
OK, let's call the block matrix above $M$. First eliminate $N$ by a substitution $c\mapsto d N$. Then substitute $z\_i \mapsto d y\_i$ to eliminate $d$. Then you can construct the Schur complement w.r.t. the first and last rows/columns of $M$ to get $\det(M) = -c^2 N x\_1^2 x\_2^2 x\_3^2 \det P$, with
$$
P = 
\begin{pmatrix}
x\_1 y\_1 + x\_1^{-1}-x\_2^{-1}-{x\_1^{-2}} & x\_1^{-1}-x\_2^{-1}+x\_2^{-2} &
 x\_1^{-1}-x\_2^{-1} \\
x\_1 y\_2+x\_2^{-1}-x\_3^{-1} & x\_2 y\_2+x\_2^{-1}-x\_3^{-1}-x\_2^{-2} &
 x\_2^{-1}-x\_3^{-1}+x\_3^{-2} \\
 x\_1 & x\_2 & x\_3 \\
\end{pmatrix}
$$
It might be easier to discuss this determinant.


