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Positive Schur complements for finite Gaussian precisions

Statement

Let $r \geq 0$ and $s \geq 1$ be integers. For $n \geq 1$ , a real matrix $M \in R^{n \times n}$ is symmetric positive definite if $M^{T} = M$ and $v^{T} M v > 0$ for every nonzero $v \in R^{n}$ . Let $A \in R^{r \times r}$ , $B \in R^{r \times s}$ , and $C \in R^{s \times s}$ , and set

$K : = (\begin{array}{cc} A & B \\ B^{T} & C \end{array}) \in R^{(r + s) \times (r + s)} .$

Assume $K$ is symmetric positive definite. Then $C$ is symmetric positive definite, $\det C > 0$ , and $C^{- 1}$ exists. The Schur complement $S : = A - B C^{- 1} B^{T}$ is symmetric positive definite when $r \geq 1$ ; when $r = 0$ , $S$ is the empty $0 \times 0$ matrix. For all $x \in R^{r}$ and $y \in R^{s}$ ,

${(\begin{array}{c} x \\ y \end{array})}^{T} K (\begin{array}{c} x \\ y \end{array}) = x^{T} S x + (y + C^{- 1} B^{T} x)^{T} C (y + C^{- 1} B^{T} x) .$

For $r = 0$ , the first term and $S$ are interpreted on the singleton space $R^{0}$ .

Statement references

A matrix admits a Cholesky factorisation with positive diagonal exactly when it is Hermitian positive definite, and that factor is unique

Facts & Assumptions

Given: A symmetric positive-definite real block matrix $K$ with block sizes $r \geq 0$ and $s \geq 1$ as in the Statement.

[F1]

Every real symmetric positive-definite $s \times s$ matrix has a factorization $C = L L^{T}$ with $L$ lower triangular and positive diagonal; the determinant of a triangular matrix is the product of its diagonal entries, determinants are multiplicative, and a real square matrix with nonzero determinant is invertible (A matrix admits a Cholesky factorisation with positive diagonal exactly when it is Hermitian positive definite, and that factor is unique, The determinant of a triangular matrix is the product of its diagonal entries, For same-sized finite square matrices over a commutative ring, $\det (A B) = \det (A) \det (B)$ , A finite square real matrix is invertible if and only if its determinant is nonzero).

Proof

1.1givenF1algebra

For nonzero $z \in R^{s}$ , the vector $(0, z)$ is nonzero, so $z^{T} C z = (0, z)^{T} K (0, z) > 0$ . Thus $C$ is symmetric positive definite. By the Cholesky statement in [F1], $C = L L^{T}$ with positive diagonal; by the determinant laws in [F1], $\det C = (\det L)^{2} = (\prod_{i} L_{i i})^{2} > 0$ , so the invertibility criterion in [F1] gives that $C$ is invertible.

2.1step 1.1algebra

Since $C$ is symmetric and invertible, $C^{- 1}$ is symmetric. Expanding the quadratic form on the right gives $x^{T} (A - B C^{- 1} B^{T}) x + 2 x^{T} B y + y^{T} C y$ , which is the quadratic form on the left; this proves the displayed completion-of-the-square identity.

3.1givenstep 1.1step 2.1constructalgebra∎

If $r \geq 1$ and $x \neq 0$ , choose $y = - C^{- 1} B^{T} x$ in step 2.1. The second term vanishes and $(x, y) \neq 0$ , so positive definiteness of $K$ gives $x^{T} S x > 0$ . Symmetry of $S$ follows from symmetry of $A, C^{- 1}$ ; hence $S$ is symmetric positive definite. If $r = 0$ , the claimed empty-matrix convention holds by definition.

Positive Schur complements for finite Gaussian precisions

Statement

Statement references

Facts & Assumptions

Proof

Depends on

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