Suppose I have a linear system $Ax = b$. The matrix $A \in \mathbb{R}^{m \times m}$ has $Rank(A) = n < m$, so it is a singular matrix. Suppose I separate the variables into three groups: $x_i, x_-$ and $x_+$.
\begin{equation*} \begin{bmatrix}A_{ii}&A_{i+}&A_{i-}\\A_{+i}&A_{++}&A_{+-}\\A_{i-}&A_{+-}&A_{--} \end{bmatrix} \begin{bmatrix}x_{i}\\x_{+}\\x_{-}\end{bmatrix} = \begin{bmatrix}b_{i}\\b_{+}\\b_{-}\end{bmatrix} \end{equation*}
Then I set some variables $x_+ = x_- $, which lead to the following reduced linear system:
\begin{equation*} \begin{bmatrix}A_{ii}&A_{i-} + A_{i+} \\ A_{i-} + A_{+i}&A_{--} + A_{-+} + A_{+-} + A_{++} \end{bmatrix} \begin{bmatrix}x_{i}\\x_{-}\end{bmatrix} = \begin{bmatrix}b_{i}\\b_{-} + b_{+}\end{bmatrix} \end{equation*}
My question is, is there any relation between the singularities of the full system and the reduced one?
I'm facing the problem of a rigid body in dynamics. The stiffness matrix is singular and you have to restrict the 6 degrees of freedom related to translation and rotation. Here my $x$ belongs to the displacement vector, so with the equation $x_+ = x_-$ I'm setting some displacements to be equal to others.
Weird thing is that my resultant system is full rank, but it does not make sense in my problem since there is an infinite number of displacements which fulfills that restriction. I expect the rank to be $l-6 \leq n \lt l$ (with $l$ number of columns in reduced $A$), but not full rank.
Thank you for your time!
Solved the problem. The full rank was because of bad assembling of the reduced matrix. Answering the question:
is there any relation between the singularities of the full system and the reduced one?
Yes, but the number of dependent rows can decrease. It depends on which variables are coupled.
