GraphMath

Schur complement

Block row reduction for determinant and inverse computation

How can block structure simplify determinant and inverse calculations?

Many matrices contain useful block structure. The Schur complement formalizes block elimination: after partitioning a square matrix into blocks, the expression S = D - CA⁻¹B is the block that remains after eliminating C using A. This chapter uses that block-level row reduction to compute determinants and inverses.

Key ideas

Schur complement is ordinary row reduction performed at the block level.

  • A square matrix is partitioned into blocks so that the diagonal block A is square and invertible
  • The Schur complement S = D - CA⁻¹B is the block left after eliminating C using A
  • The block row operation is performed by left-multiplying with a block lower triangular matrix
  • For determinant computation, elimination produces an upper block triangular matrix, so |M| = |A|×|S|
  • For inverse computation, the same block elimination continues to reduced echelon form
  • If A is not invertible, the chapter explains why one may need to start from the other diagonal block instead

The main point is not a new kind of operation, but the same elimination logic applied to structured blocks instead of individual entries.

Why does the expression D - CA⁻¹B appear?

It appears when the lower-left block C is eliminated using the invertible block A as a pivot. The same row-replacement logic used in ordinary elimination subtracts CA⁻¹ times the top block row from the bottom block row, leaving D - CA⁻¹B in the lower-right position.

Related chapters

Chapter contents

The PDF is a single document. The page links below are best-effort: most browsers support them, but some viewers may ignore the page hint.

Topic Pages
Block partition of square matrix 1–2
Definition of Schur complement 2–3
Schur complement for determinant computation 3–4
Schur complement for M⁻¹ computation 4–6
Summary 6–7

Why does Schur complement depend on the chosen pivot block?

Because the formula uses A⁻¹, so A must be invertible to serve as the block pivot. If that pivot block is not invertible, the same elimination idea may have to start from the other diagonal block instead, just as ordinary row reduction may need a different pivot when the current pivot is zero.

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