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Month: August 2010

Schur complement and geometry of positive definite matrices

This post concerns the geometry of the convex cone of positive definite symmetric matrices. Let us denote by Sn the set of n×n real symmetric matrices, and by S+n the subset of positive definite matrices. Let us fix some KSn and write

K:=(AWWS)

with ASnm and SSm for some fixed 1m<n. If AS+nm then the Schur complement of A with respect to K is the matrix

T:=SWA1WSm.

It is well known that

KS+niffAS+nmandTS+m,

and moreover, K1 admits the block decomposition

K1=(InA1W0Im)(A100T1)(In0WA1Im),

see for instance Theorem 7.7.6 page 472 in Horn and Johnson, Matrix analysis, Cambridge University Press, 1990. In terms of Gaussian random vectors, if X=(Y,Z) is a Gaussian random vector of Rn=Rnm×Rm with 1m<n, with covariance matrix K, then

YN(0,A)andZN(0,S)andL(Z|Y)=N(WA1Y,T).

More generally, for a non empty I{1,,n}, we have, with J:=Ic,

XIN(0,KI,I)andXJN(0,KJ,J)

and the Schur complement of KJ,J in K is KI,IKI,J(KJ,J)1KJ,I and

L(XI|XJ)N(KI,J(KJ,J)1XJ,KI,IKI,J(KJ,J)1KJ,I).

Moreover, from the block inverse formula we get

(K1)I,I=(KI,IKI,J(KJ,J)1KJ,I)1.

The components of XI are conditionally independent given the remaining components iff ((K1)I,I)1 is diagonal, i.e. iff (K1)I,I is diagonal. In particular, if I={i,j} with ij then Xi and Xj are conditionally independent given the remaining components iif (K1)i,j=0.

The extremal elements of the closed convex cone ¯S+n are the unit rank matrices {uu:uRn}. Every A¯S+n can be written as A=λ1u1u1++λnunun where λ1,,λn and u1,,un denote the spectrum and the eigenvectors of A respectively. This is the Carathéodory local parametrization of the convex cone ¯S+n with n Carathéodory conic rays λ1u1u1,,λrunun.

For any KSn, any 1ijn, and any xR, let ξi,j(K,x) be the element of Sn obtained from K by replacing Ki,j and Kj,i by x. In particular,

K=ξi,j(K,Ki,j).

Now for which values of x the matrix ξi,j(K,x) belongs to S+n? The answers is as follows.

Theorem 1 Set KS+n, n2, 1ijn, and xR.

  1. Off-diagonal. If i<j then

    ξi,j(K,x)S+niffδi,j(K)<x<δ+i,j(K)

    where δ±1,2(K):=±K1,1K2,2 if n=2, while if n>2,

    δ±i,j(K):=uv±((uv)2(A1)i,i(vBvKj,j))1/2(A1)i,i.

    with

    A:=Kj,j,B:=(A1)i,i,u:=Ai,i,v:=K{i,j},j.

  2. Diagonal.

    ξi,i(K,x)S+niffx>δi,i(K),

    where

    δi,i(K):=wW1w

    with

    W:=Ki,iw:=Ki,i.

In particular, δi,i(K)<Ki,i for any 1in and δi,j(K)<Ki,j<δ+i,j(K) for any 1i<jn. Moreover, ξi,j(K,0)S+n iff vBvKj,j.

Proof: We prove the first part for n>2 and leave the remaining statements to the reader. The matrix A is non singular since K is non singular. Moreover, KS+n implies A1S+n, and in particular (A1)i,i>0. The set

{xR:ξi,j(K,x)S+n}

is non empty and convex (an interval), as the intersection of an affine subspace with a convex cone, both containing K=ξi,j(K,Ki,j). For any xR,

(ξi,j(K,x))j,j=Kj,j=A.

By considering the Schur complement, the matrix ξi,j(K,x) belongs to S+n iff

1k,knzk(A1)k,kzk<Kj,j,

where z:=(ξi,j(K,x))j,j. By rewriting z in terms of K, we get the equation

(A1)i,ix2+(2uv)x+vBvKj,j<0.

Recall that (A1)i,i>0. Next, since K=ξi,j(K,Ki,j) belongs to S+n, the equation admits x=Ki,j as a solution, and thus the discriminant

(uv)2(A1)i,i(vBvKj,j)

is positive. This leads to the non empty interval (δi,j(K),δ+i,j(K)). ☐

Let KS+n with n>2, and consider the notations of theorem 1. For any 1i<jn, we define the matrices ξi,j(K) and ξ+i,j(K) by

ξ±i,j(K):=ξi,j(K,δ±i,j(K)).

Similarly, for any 1in, we define

ξi,i(K):=ξi,i(K,δi,i(K)).

The elements of

E(K):={ξ±i,j(K);1i<jn}{ξi,i(K);1in}

are not of full rank and belong to the boundary of S+n. If E1,,ErE(K), and λ1>0,,λr>0, then λ1E1++λrErS+n. These sums construct from K an open convex cone, included in S+n, and containing K.

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