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 K∈Sn and write
K:=(AWW⊤S)
with A∈Sn−m and S∈Sm for some fixed 1≤m<n. If A∈S+n−m then the Schur complement of A with respect to K is the matrix
T:=S−W⊤A−1W∈Sm.
It is well known that
K∈S+niffA∈S+n−mandT∈S+m,
and moreover, K−1 admits the block decomposition
K−1=(In−A−1W0Im)(A−100T−1)(In0−W⊤A−1Im),
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=Rn−m×Rm with 1≤m<n, with covariance matrix K, then
Y∼N(0,A)andZ∼N(0,S)andL(Z|Y)=N(W⊤A−1Y,T).
More generally, for a non empty I⊂{1,…,n}, we have, with J:=Ic,
XI∼N(0,KI,I)andXJ∼N(0,KJ,J)
and the Schur complement of KJ,J in K is KI,I−KI,J(KJ,J)−1KJ,I and
L(XI|XJ)∼N(KI,J(KJ,J)−1XJ,KI,I−KI,J(KJ,J)−1KJ,I).
Moreover, from the block inverse formula we get
(K−1)I,I=(KI,I−KI,J(KJ,J)−1KJ,I)−1.
The components of XI are conditionally independent given the remaining components iff ((K−1)I,I)−1 is diagonal, i.e. iff (K−1)I,I is diagonal. In particular, if I={i,j} with i≠j then Xi and Xj are conditionally independent given the remaining components iif (K−1)i,j=0.
The extremal elements of the closed convex cone ¯S+n are the unit rank matrices {uu⊤:u∈Rn}. Every A∈¯S+n can be written as A=λ1u1u⊤1+⋯+λnunu⊤n 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 λ1u1u⊤1,…,λrunu⊤n.
For any K∈Sn, any 1≤i≤j≤n, and any x∈R, 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 K∈S+n, n≥2, 1≤i≤j≤n, and x∈R.
- Off-diagonal. If i<j then
ξi,j(K,x)∈S+niffδ−i,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):=−u⊤v±((u⊤v)2−(A−1)i,i(v⊤Bv−Kj,j))1/2(A−1)i,i.
with
A:=K−j,−j,B:=(A−1)−i,−i,u:=A−i,i,v:=K−{i,j},j.
- Diagonal.
ξi,i(K,x)∈S+niffx>δi,i(K),
where
δi,i(K):=w⊤W−1w
with
W:=K−i,−iw:=K−i,i.
In particular, δi,i(K)<Ki,i for any 1≤i≤n and δ−i,j(K)<Ki,j<δ+i,j(K) for any 1≤i<j≤n. Moreover, ξi,j(K,0)∈S+n iff v⊤Bv≤Kj,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, K∈S+n implies A−1∈S+n, and in particular (A−1)i,i>0. The set
{x∈R: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 x∈R,
(ξi,j(K,x))−j,−j=K−j,−j=A.
By considering the Schur complement, the matrix ξi,j(K,x) belongs to S+n iff
∑1≤k,k′≤nzk(A−1)k,k′zk′<Kj,j,
where z:=(ξi,j(K,x))−j,j. By rewriting z in terms of K, we get the equation
(A−1)i,ix2+(2u⊤v)x+v⊤Bv−Kj,j<0.
Recall that (A−1)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
(u⊤v)2−(A−1)i,i(v⊤Bv−Kj,j)
is positive. This leads to the non empty interval (δ−i,j(K),δ+i,j(K)). ☐
Let K∈S+n with n>2, and consider the notations of theorem 1. For any 1≤i<j≤n, we define the matrices ξ−i,j(K) and ξ+i,j(K) by
ξ±i,j(K):=ξi,j(K,δ±i,j(K)).
Similarly, for any 1≤i≤n, we define
ξi,i(K):=ξi,i(K,δi,i(K)).
The elements of
E(K):={ξ±i,j(K);1≤i<j≤n}∪{ξi,i(K);1≤i≤n}
are not of full rank and belong to the boundary of S+n. If E1,…,Er∈E(K), and λ1>0,…,λr>0, then λ1E1+⋯+λrEr∈S+n. These sums construct from K an open convex cone, included in S+n, and containing K.
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