$P, Q$ are symmetric square. If $P \geq Q > 0$ then show $P^{-1} \leq Q^{-1}$

Question

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inequalty concering inverses of positive definite matrix
Is $B^{-1}-A^{-1}$ a positive definite matrix?

$P, Q$ are symmetric square. If $P \geq Q > 0$ then show $P^{-1} \leq Q^{-1}$.

Thanks for your help!

Answer 1

We can prove this using the fact that if $A$ and $B$ are real symmetric matrices and $A \geq B$, then $M^T A M \geq M^T B M$ for any $M$. We'll also use the fact that if $I \geq A > 0$, then $I \leq A^{-1}$.

Let $P = L L^T$ be a Cholesky factorization of $P$. Then \begin{align*} & P \geq Q > 0 \\ \implies& L L^T \geq Q > 0 \\ \implies& I \geq L^{-1} Q L^{-T} > 0\\ \implies& I \leq L^T Q^{-1} L \\ \implies& L^{-T} L^{-1} \leq Q^{-1} \\ \implies& P^{-1} \leq Q^{-1} . \end{align*}