I am solving a problem in which, the product of the transpose of a sparse matrix $M^T$ and the matrix $M$ is a large, sparse square matrix having the form:
$$ \begin{array}{ccccccccc} 1\; & 2\; & 3\; & 4\; & 5\; & 6\; & 7\; & 8\; & 9 \end{array} $$
$$ \begin{array}{c} 1 \\ 2 \\ 3 \\ 4 \\ 5 \\ 6 \\ 7 \\ 8 \\ 9 \\ \end{array} \left[\begin{array}{c:c:c:c:c:c:c:c:c} \\ \hline \hspace{2mm} & A & \hspace{2mm} & B\\ \hline \\ \hline & C & & D & \hspace{2mm} & E\\ \hline \\ \hline & & & F & & G & \hspace{2mm} & H \\ \hline \\ \hline & & & & & I & & J & \hspace{2mm}\\ \hline \\ \end{array}\right] $$
where $A \ldots J$ are dense 3 $\times$ 3 matrices and the odd numbered blocks of rows and columns have the same "height" and "width" respectively. Although this matrix is very large the zero blocks form the majority of the matrix. In fact in this example there are only 12 non-zero rows and columns.
This product is the on the LHS of a linear system
$$ M^T M x = b $$
that I'd like to solve. Where the rows of the RHS vector are partitioned into zero and non-zero block-vectors of the same height as the row blocks of LHS matrix, i.e.
$$ b = \left[\begin{array}{c:c:c:c:c:c:c:c:c} \hspace{1mm} & b_2 & \hspace{1mm} & b_4 & \hspace{1mm} & b_6 & \hspace{1mm} & b_8 & \hspace{1mm} \end{array}\right]^T $$
And what I'd like to know is whether the solution of the original linear system is in any way related to the solution of the reduced (12 $\times$ 12) system:
$$ \left[\begin{array}{cccc} A & B\\ C & D & E\\ & F & G & H \\ & & I & J \hspace{2mm}\\ \end{array}\right] \left[\begin{array}{c} x_2 \\ x_4 \\ x_6 \\ x_8 \\ \end{array}\right] = \left[\begin{array}{c} b_2 \\ b_4 \\ b_6 \\ b_8 \\ \end{array}\right] $$
Yes.
Suppose $(a_2,a_4, a_6, a_8)$ is a solution of the latter system. Then, $(x, a_2,y, a_4,z, a_6,w, a_8)$ is a solution of the former system for all $x,y,z,w \in \mathbb{R}^3$. That's because in any equation these will get multiplied by zero.
You can safely ignore rows will all zeros since the equation $\sum 0x_i = 0$ is trivial.