Solving sub-matrix preserving overall linear system

Question

I have a linear system

$$ \begin{pmatrix} A_{11} \ A_{12} \ A_{13}\\ A_{21} \ A_{22} \ A_{23}\\ A_{31} \ A_{32} \ A_{33}\\ \end{pmatrix} \cdot \begin{pmatrix} x_1 \\ x_2 \\ x_3 \end{pmatrix} = \begin{pmatrix} b_1 \\ b_2 \\ b_3 \end{pmatrix} $$

which comes from the discretization of a system of PDEs. The overall system is large, non-symmetric and contains a saddle point problem.

Let's assume I know something about the sub-matrix $A_{2,2}$, which I can use to solve the sub-matrix more efficient. For example $A_{2,2}$ might be a diagonal matrix or symmetric or I have a suitable pre-conditioner for it.

What can I do with this knowledge. Can I precondition or even solve $A_{2,2}$ (replacing $A_{2,2}$ by $I$ and adjusting $b_2$)? How does this affect the other parts of the matrix or the right hand side?

Edit: Additional ideas: $A_{ij}$ for $i\ne j$ (non-diagonal blocks) are very sparse and are only for coupling the systems of equations. Can there be made a connection to Block Jacobi methods / preconditioning which kind of tries the same?

Answer 1

The most basic preconditioner for $$A= \begin{pmatrix} A_{11} \ A_{12} \ A_{13}\\ A_{21} \ A_{22} \ A_{23}\\ A_{31} \ A_{32} \ A_{33}\\ \end{pmatrix}$$ is a Jacobi (or diagonal) preconditioner, i.e. $\mathrm{diag}(A)$. Now if you have suitable pre-conditioner for $A_{22}$, say $P_{22}$, then I would try to use the following preconditioner $$P= \begin{pmatrix} \mathrm{diag}(A_{11}) & 0 & 0\\ 0 &P_{22} & 0\\ 0 & 0 & \mathrm{diag}( A_{33})\\ \end{pmatrix}$$

If your matrix is block n-diagonal, then I would suggest to read the following thing. I didn't read it all, the name is block-matrix preconditioning, but at first pages looks like they talking about a block 3-diagonal. Perhaps later they generalize it. For a PDE discretization you get such matrices, so hopefully this is what you need.

Answer 2

I know I'm giving you bad news, but what you in fact have is this (where obviously $A_{jk}$ are block matrices): $$A_{11}x_{1}+A_{12}x_{2}+A_{13}x_{3}=b_{1}$$ $$A_{21}x_{1}+A_{22}x_{2}+A_{23}x_{3}=b_{2}$$ $$A_{31}x_{1}+A_{32}x_{2}+A_{33}x_{3}=b_{3}$$ Now...unfortunately if you don't have any other restriction I guess that's as far as you can get. If you other symmetries maybe you can go further. The fact that you solved the block matrix $A_{22}=b_2$ unfortunately doesn't help you much. Indeed is more or less like knowing that $y=7$ doesn't help you in find $x,z$ in the equation $$x+y+z=7$$ I'm sorry but I hope this answer was anyway useful telling to look other directions...