---
tag: R997ZAD141
kind: statement
role: proposition
name: Schur complement method
msc: [65N55, 65F05, 65F10]
uses: [6HWFBXYPZP, 0ASDFEQS6P, 0KX66CZMBB]
about: [AN054RHT1K, 6HWFBXYPZP]
sources:
  - url: https://en.wikipedia.org/wiki/Schur_complement_method
same_as:
  wikidata: Q17103137
  wikipedia: Schur complement method
---

## Statement

Let $A$ be an invertible real matrix.
Write it in [[block form|0ASDFEQS6P]] as
$$A = \begin{pmatrix} A_{II} & A_{I\Gamma} \\ A_{\Gamma I} & A_{\Gamma\Gamma} \end{pmatrix}, \qquad A_{II} = \begin{pmatrix} A_{11} & & \\ & \ddots & \\ & & A_{NN} \end{pmatrix}.$$
Here the unknowns are split into interior unknowns $u_I = (u_1, \dots, u_N)$ of $N$ subdomains and interface unknowns $u_\Gamma$.
Each block $A_{jj}$ is invertible.
Let $S = A_{\Gamma\Gamma} - A_{\Gamma I} A_{II}^{-1} A_{I\Gamma}$ be the [[Schur complement|6HWFBXYPZP]] of $A_{II}$ in $A$.
Then $(u_I, u_\Gamma)$ solves $A u = f$ with $f = (f_I, f_\Gamma)$ if and only if
$$S u_\Gamma = f_\Gamma - A_{\Gamma I} A_{II}^{-1} f_I \quad \text{and} \quad u_j = A_{jj}^{-1} \bigl( f_j - (A_{I\Gamma} u_\Gamma)_j \bigr) \text{ for } j = 1, \dots, N.$$
If $A$ is symmetric [[positive definite|0KX66CZMBB]], then $S$ is symmetric positive definite.

## Remarks

The equivalence is block Gaussian elimination: the first block row gives $u_I = A_{II}^{-1}(f_I - A_{I\Gamma} u_\Gamma)$, and substituting into the second block row gives the interface equation.
Since $A_{II}$ is block diagonal, every application of $A_{II}^{-1}$ splits into independent solves on the subdomains.
If $A$ is symmetric positive definite, then $v^T S v = w^T A w$ for $w = (-A_{II}^{-1} A_{I\Gamma} v, v)$, which is positive for $v \ne 0$.

In practice one does not form $S$.
One solves the interface system by the [[conjugate gradient method|AFXZGQB4EN]], and each product $S v$ costs one solve with each $A_{jj}$.
Take a finite element discretization of an elliptic problem with a fixed set of subdomains.
Then the condition number of $S$ grows like $h^{-1}$ in the mesh size $h$, against $h^{-2}$ for $A$.
Preconditioners for $S$, such as [[balancing domain decomposition|X8MCVW56QP]], reduce this further.

## Explanations

### L1 human

Cut the domain of a problem into pieces that touch only along thin interfaces.
If the values on the interfaces were known, each piece could be solved on its own.
The method first finds the interface values from a smaller system, the Schur complement system, and then fills in each piece independently.

### L2 human

This is the non-overlapping branch of [[domain decomposition methods|AN054RHT1K]], also called iterative substructuring.
The Schur complement acts on interface functions like a discrete Steklov–Poincaré operator, mapping Dirichlet data on the interfaces to the jump of the normal fluxes.
Its solution by a preconditioned Krylov method is the starting point of the Neumann–Neumann, BDD and FETI families.