Inequality between two matrices

Question

Given a full rank $n \times m$ matrix $K$ with $m<n$ and an invertible symmetric matrix $J$. Let $A$ be a symmetric positive semi-definite $n \times n$ matrix such that \begin{equation} (K^T K)^{-1}K^TAK(K^TK)^{-1}-(K^TJK)^{-1} \end{equation} is positive semi-definite. I'd like to prove that \begin{equation} A-K(K^TJK)^{-1}K^T \end{equation} is also positive semi-definite.

This is my investigation: \begin{align*} (K^T K)^{-1}K^TAK(K^TK)^{-1}\geq(K^TJK)^{-1}\iff K^TAK\geq K^T K(K^TJK)^{-1}K^T K \end{align*} by multiplying RHS by $K^T K$ from left and right. But, I am not sure if we can multiply bu such matrices and I don't know what I do then. Could anyone help me? Thanks in advance.

Answer 1

Well you can get from \begin{equation} (K^T K)^{-1}K^TAK(K^TK)^{-1}-(K^TJK)^{-1} \end{equation} to \begin{equation} A-K(K^TJK)^{-1}K^T \end{equation} by left-multiplying by $K=(K^T)^{-1}K^T K$ and right-multiplying by $K^T = K^T K K^{-1}$. So all you need to do is show that doing this will preserve positivity. This follows because a symmetric positive semidefinite matrix can always be decomposed as $M=B B^T$ (for the real case that I assume you're dealing with here, given that we have transposes rather than hermitian conjugates), and conversely any matrix with such a decomposition is positive. So for positive semidefinite $M$: \begin{equation} K M K^T = K (B B^T)K^T \geq 0. \end{equation}

See, for example, http://www.math.ucsd.edu/~njw/Teaching/Math271C/Lecture_03.pdf