This post concerns the geometry of the convex cone of positive definite symmetric matrices. Let us denote by ${\mathcal{S}_n}$ the set of ${n\times n}$ real symmetric matrices, and by ${\mathcal{S}_n^+}$ the subset of positive definite matrices. Let us fix some ${K\in\mathcal{S}_n}$ and write

$K:= \begin{pmatrix} A & W \\ W^\top & S \end{pmatrix}$

with ${A\in\mathcal{S}_{n-m}}$ and ${S\in\mathcal{S}_{m}}$ for some fixed ${1\leq m<n}$. If ${A\in\mathcal{S}_{n-m}^+}$ then the Schur complement of ${A}$ with respect to ${K}$ is the matrix

$T:=S-W^\top A^{-1}W\in\mathcal{S}_m.$

It is well known that

$K\in\mathcal{S}_n^+ \quad\text{iff}\quad A\in\mathcal{S}_{n-m}^+ \quad\text{and}\quad T\in\mathcal{S}_m^+,$

and moreover, ${K^{-1}}$ admits the block decomposition

$K^{-1} = \begin{pmatrix} I_n & -A^{-1}W \\ 0 & I_m \\ \end{pmatrix} \begin{pmatrix} A^{-1} & 0 \\ 0 & T^{-1} \\ \end{pmatrix} \begin{pmatrix} I_n & 0 \\ -W^\top A^{-1} & I_m \\ \end{pmatrix},$

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 ${\mathbb{R}^n=\mathbb{R}^{n-m}\times\mathbb{R}^m}$ with ${1\leq m<n}$, with covariance matrix ${K}$, then

$Y\sim\mathcal{N}(0,A) \quad\text{and}\quad Z\sim\mathcal{N}(0,S) \quad\text{and}\quad \mathcal{L}(Z\,\vert\,Y)=\mathcal{N}(W^\top A^{-1}Y,T).$

More generally, for a non empty ${I\subset\{1,\ldots,n\}}$, we have, with ${J:=I^c}$,

$X_I\sim\mathcal{N}(0,K_{I,I}) \quad\text{and}\quad X_{J}\sim\mathcal{N}(0,K_{J,J})$

and the Schur complement of ${K_{J,J}}$ in ${K}$ is ${K_{I,I}-K_{I,J}(K_{J,J})^{-1}K_{J,I}}$ and

$\mathcal{L}(X_I\,\vert\,X_{J}) \sim\mathcal{N}(K_{I,J}(K_{J,J})^{-1}X_{J},\,K_{I,I}-K_{I,J}(K_{J,J})^{-1}K_{J,I}).$

Moreover, from the block inverse formula we get

$(K^{-1})_{I,I} = (K_{I,I}-K_{I,J}(K_{J,J})^{-1}K_{J,I})^{-1}.$

The components of ${X_I}$ 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 ${i\neq j}$ then ${X_i}$ and ${X_j}$ are conditionally independent given the remaining components iif ${(K^{-1})_{i,j}=0}$.

The extremal elements of the closed convex cone ${\overline{\mathcal{S}_n^+}}$ are the unit rank matrices ${\{uu^\top: u\in\mathbb{R}^n\}}$. Every ${A\in\overline{\mathcal{S}_n^+}}$ can be written as ${A=\lambda_1 u_1u_1^\top+\cdots+\lambda_n u_nu_n^\top}$ where ${\lambda_1,\ldots,\lambda_n}$ and ${u_1,\ldots,u_n}$ denote the spectrum and the eigenvectors of ${A}$ respectively. This is the Carathéodory local parametrization of the convex cone ${\overline{\mathcal{S}_n^+}}$ with ${n}$ Carathéodory conic rays ${\lambda_1 u_1u_1^\top,\ldots,\lambda_r u_nu_n^\top}$.

For any ${K\in\mathcal{S}_n}$, any ${1\leq i\leq j\leq n}$, and any ${x\in\mathbb{R}}$, let ${\xi_{i,j}(K,x)}$ be the element of ${\mathcal{S}_n}$ obtained from ${K}$ by replacing ${K_{i,j}}$ and ${K_{j,i}}$ by ${x}$. In particular,

$K=\xi_{i,j}(K,K_{i,j}).$

Now for which values of ${x}$ the matrix ${\xi_{i,j}(K,x)}$ belongs to ${\mathcal{S}_n^+}$? The answers is as follows.

Theorem 1 Set ${K\in\mathcal{S}_n^+}$, ${n\geq2}$, ${1\leq i\leq j\leq n}$, and ${x\in\mathbb{R}}$.

1. Off-diagonal. If ${i<j}$ then

$\xi_{i,j}(K,x)\in\mathcal{S}_n^+ \quad\text{iff}\quad \delta_{i,j}^-(K)<x<\delta_{i,j}^+(K)$

where ${\delta_{1,2}^{\pm}(K):=\pm\sqrt{K_{1,1}K_{2,2}}}$ if ${n=2}$, while if ${n>2}$,

$\delta_{i,j}^\pm(K):= \frac{-u^\top v\pm\left((u^\top v)^2- (A^{-1})_{i,i}(v^\top Bv-K_{j,j})\right)^{1/2}} {(A^{-1})_{i,i}}.$

with

$A:=K_{-j,-j},\quad B:=(A^{-1})_{-i,-i},\quad u:=A_{-i,i},\quad v:=K_{-\{i,j\},j}.$

2. Diagonal.

$\xi_{i,i}(K,x)\in\mathcal{S}_n^+ \quad\text{iff}\quad x > \delta_{i,i}(K),$

where

$\delta_{i,i}(K):=w^\top W^{-1}w$

with

$W:=K_{-i,-i}\quad w:=K_{-i,i}.$

In particular, ${\delta_{i,i}(K)<K_{i,i}}$ for any ${1\leq i\leq n}$ and ${\delta_{i,j}^-(K)< K_{i,j}<\delta_{i,j}^+(K)}$ for any ${1\leq i<j\leq n}$. Moreover, ${\xi_{i,j}(K,0)\in\mathcal{S}_n^+}$ iff ${v^\top Bv\leq K_{j,j}}$.

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, ${K\in\mathcal{S}_n^+}$ implies ${A^{-1}\in\mathcal{S}_n^+}$, and in particular ${(A^{-1})_{i,i}>0}$. The set

$\{x\in\mathbb{R}:\xi_{i,j}(K,x)\in\mathcal{S}_n^+\}$

is non empty and convex (an interval), as the intersection of an affine subspace with a convex cone, both containing ${K=\xi_{i,j}(K,K_{i,j})}$. For any ${x\in\mathbb{R}}$,

$(\xi_{i,j}(K,x))_{-j,-j}=K_{-j,-j}=A.$

By considering the Schur complement, the matrix ${\xi_{i,j}(K,x)}$ belongs to ${\mathcal{S}_n^+}$ iff

$\sum_{1\leq k,k’\leq n} z_k (A^{-1})_{k,k’} z_{k’} <K_{j,j},$

where ${z:=(\xi_{i,j}(K,x))_{-j,j}}$. By rewriting ${z}$ in terms of ${K}$, we get the equation

$(A^{-1})_{i,i}\,x^2+(2u^\top v)\,x+v^\top Bv-K_{j,j}<0.$

Recall that ${(A^{-1})_{i,i}>0}$. Next, since ${K=\xi_{i,j}(K,K_{i,j})}$ belongs to ${\mathcal{S}_n^+}$, the equation admits ${x=K_{i,j}}$ as a solution, and thus the discriminant

$(u^\top v)^2- (A^{-1})_{i,i}(v^\top Bv-K_{j,j})$

is positive. This leads to the non empty interval ${(\delta_{i,j}^-(K),\delta_{i,j}^+(K))}$. ☐

Let ${K\in\mathcal{S}_n^+}$ with ${n>2}$, and consider the notations of theorem 1. For any ${1\leq i<j\leq n}$, we define the matrices ${\xi_{i,j}^-(K)}$ and ${\xi_{i,j}^+(K)}$ by

$\xi_{i,j}^\pm(K):=\xi_{i,j}(K,\delta_{i,j}^\pm(K)).$

Similarly, for any ${1\leq i\leq n}$, we define

$\xi_{i,i}(K):=\xi_{i,i}(K,\delta_{i,i}(K)).$

The elements of

$\mathcal{E}(K):= \{\xi_{i,j}^\pm(K);1\leq i< j\leq n\} \cup\{\xi_{i,i}(K);1\leq i\leq n\}$

are not of full rank and belong to the boundary of ${\mathcal{S}_n^+}$. If ${E_1,\ldots,E_r\in\mathcal{E}(K)}$, and ${\lambda_1>0,\ldots,\lambda_r>0}$, then ${\lambda_1 E_1+\cdots+\lambda_r E_r\in\mathcal{S}_n^+}$. These sums construct from ${K}$ an open convex cone, included in ${\mathcal{S}_n^+}$, and containing ${K}$.

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