If Σ1≤Σ2, then Σ−11≥Σ−12
Note that by assumption B=Σ2−Σ1 is positive semi-definite. If I can show that the resolvent identity of Σ−11−Σ−12, Σ−12(Σ2−Σ1)Σ−11 is positive semi-definite, then the above statement is verified.
This requires the following two results:
Sylvester's Law of Inertia
Symmetric matrices A and B are congruent (i.e. there is a non-singular matrix C such that CTAC=B) if and only if A and B have the same inertia. They have the same number of positive, negative and zero eigenvalues.Theorem 1.
The product of a symmetric positive definite matrix A and a symmetric matrix B has the same inertia as BProof
Note that A−1/2ABA1/2=A1/2BA1/2. The right hand side is similar to AB, which means they have the same eigenvalues. Since A1/2 is symmetric, the matrix A1/2BA1/2 is congruent to B. By Sylvester's Law of inertia, the eignevalues of B have the same inertia as A1/2BA1/2 and also of AB.
The main point from Theorem 1 is that multiplying a positive definite matrix A to any symmetric matrix B will not change the inertia of the result. Note that Σ−12 and Σ−11 are positive definite. We let Σ−12=A1, Σ−11=A2 and Σ2−Σ1=B, and applying Theorem 1 twice, leads to Σ−12(Σ2−Σ1)Σ−11 positive semi-definite. We have shown Σ−11−Σ−12 is indeed positive semi-definite.
Caution!
For real positive definite matrices, not all of them are symmetric! For example, given a symmetric positive definite matrix B and a anti-symmetric matrix C (CT=−C), the sum of which (A=B+C) is positive definite. Extra care must be taken. All references of positive definiteness are within the context of symmetric matrices.
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