$A \preceq B => B^{-1} \preceq A^{-1}$

Question

$A \preceq B$ indicates for covariance matrices A and B, B-A is SPSD (symmetric positive semi definite). Hence, I want to show that $B^{-1} \preceq A^{-1}$ , which seems straight-forward but I can't come up with a way to solve this.

Answer 1

Let $A^{1/2}$ be the unique SPSD square root of $A$, and similarly for $B$ and the inverses $A^{-1/2}$ and $B^{-1/2}$.
Then $A \preceq B$ is equivalent to $I = A^{-1/2} A A^{-1/2} \preceq A^{-1/2} B A^{-1/2} = (B^{1/2} A^{-1/2})^T (B^{1/2} A^{-1/2})$. Since $CD$ and $DC$ have the same nonzero eigenvalues, this is equivalent to $I \preceq (B^{1/2} A^{-1/2})(B^{1/2} A^{-1/2})^T = B^{1/2} A^{-1} B^{1/2}$ and thus to $B^{-1} = B^{-1/2} I B^{-1/2} \preceq B^{-1/2} B^{1/2} A^{-1} B^{1/2} B^{-1/2} = A^{-1}$.