Full-rank matrix

Question

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:

$$\left[\begin{array}{ccc} A & B & 0 \\ 0 & C & D \\ E & 0 & F\end{array}\right]=\left[\begin{array}{ccc|ccc|ccc} N & 0 & 0 & x_{1}-1 & x_{1} & x_{1} & 0 & 0 & 0\\ 0 & N & 0 & x_{2} &x_2-1 & x_{2} & 0 & 0 & 0\\ 0 & 0 & N & x_{3} & x_{3} & x_3-1 & 0 & 0 & 0\\ \hline 0 & 0 & 0 & x_1z_1 & 0 & 0 & -1 & 1 & 0\\ 0 & 0 & 0 & x_1z_2 & x_2z_2 & 0 & 0 & -1 & 1\\ 0 & 0 & 0 & x_1 & x_2 & x_3 & 0 & 0 & 0\\ \hline c & 0 & 0 & 0 & 0 & 0 & -x_{1}^2 & 0 & 0\\ 0 & c & 0 & 0 & 0 & 0 & 0 & -x_{2}^2 & 0\\ 0 & 0 & c & 0 & 0 & 0 & 0 & 0 & -x_{3}^2 \end{array}\right]$$

Answer 1

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.

Answer 2

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.