Suppose $X\in\mathbf{R}^{n\times p}$ is a real matrix with full column rank($n\geq p$), $B \in \mathbf{R}^{n\times n}$ is a positive definite matrix, i.e all eigenvalue of matrix $B$ is positive, I want to show following matrix is still positive semi-definite. $$B^{-1}-X(X^TBX)^{-1}X^T $$ I attack this problem in two step.
First I assume $n=p$ which means that X is an invertible matrix, then $B^{-1}-X(X^TBX)^{-1}X^T =B^{-1}-B^{-1}=\mathbf{0}$ is a positive semi-definite matrix.
Second I want to show when $n>p$, I am not able to show this but I do some simulations. I just let $X_{ij}\sim N(0,1) $ randomly and let $B$ be any positive definite matrix, then I find all eigenvalue of $B^{-1}-X(X^TBX)^{-1}X^T$ is greater or equal to zero, which implies positive-definiteness.
Can anybody give some hint on my second part, I'd like to get some theoretical proof. Really appreciated any suggestions and help!!
[new edited] I realized I make a mistake in the statement of question, it should be $B^{-1}-X(X^TBX)^{-1}X^T$ instead of $B-X(X^TBX)^{-1}X^T$
Since B is positive definite matrix, we know from matrix eigen-decomposition, $$B=C^TC=C^2$$ where $C$ is "square root" of $B$, which is also positive definite . To show $B^{-1}-X\left(X^T B X\right)^{-1} X^T$ is positive semi-definite, it suffices to show that $C[B^{-1}-X\left(X^T B X\right)^{-1} X^T]C^T$ is positive semi-definite. Define $Q=CX\in\mathbf{R}^{n\times p}$ $$ \begin{aligned} C[B^{-1}-X\left(X^T B X\right)^{-1} X^T]C^T&=CB^{-1}C-CX(X^TC^TCX)^{-1}X^TC^T\\ &=I_n-Q(Q^TQ)^{-1}Q^T \end{aligned} $$ since rank of Q is $p\leq n$, $Q(Q^TQ)^{-1}Q^T$ has $n-p$ eigenvalue equal to $0$, other eigenvalue equal to 1. So $I_n-Q(Q^TQ)^{-1}Q^T$ is positive semi-definite.