# How to prove positive (semi-)definitness in matrix notation without numbers

I'd like to show that $$V[\hat\beta_{OLS}]-V[\hat\beta_{GLS}]=\sigma^2(X'X)^{-1}(X'\Omega X)(X'X)^{-1}-\sigma^2(X'\Omega X)^{-1}\geq 0$$ is positive (semi-) definite. $$\Omega$$ and $$X$$ are square-matrices. I know how to check for positive (semi-) definiteness with eigenvalues if the matrices contain numbers, but I wonder if there is a quick way to check if the above expression is positive (semi-) definite when we only have matrix notation without numbers?

Note that $$X$$ is generally not square, unless you have as many regressors as observations. In that case - see the derivation at the end of my answer - OLS and GLS are equally efficient.
Notice also a mistake in the above expression in that the correct variance of GLS is $$\sigma^2(X'\Omega^{-1} X)^{-1}$$
A standard result in matrix algebra tells out that $$A\geq B\Leftrightarrow B^{-1}-A^{-1}\geq0$$ (much like $$3>2$$ and $$1/2>1/3$$). Hence, we may also establish that ($$\sigma^2$$ is just a scale factor which cancels out of the comparison) $$X'\Omega^{-1} X-X'X(X'\Omega X)^{-1}X'X\geq 0$$ Rearrange the lhs to $$X'(\Omega^{-1}-X(X'\Omega X)^{-1}X')X$$ or, with $$\Omega^{1/2}$$ a symmetric matrix square root of $$\Omega$$ and $$\Omega^{-1/2}$$ its inverse, $$X'(\Omega^{-1/2}\Omega^{-1/2}-\Omega^{-1/2}\Omega^{1/2}X(X'\Omega X)^{-1}X'\Omega^{1/2}\Omega^{-1/2})X$$ or $$X'\Omega^{-1/2}(I-\Omega^{1/2}X(X'\Omega X)^{-1}X'\Omega^{1/2})\Omega^{-1/2}X$$ or $$X'\Omega^{-1/2}(I-\Omega^{1/2}X(X'\Omega^{1/2}\Omega^{1/2} X)^{-1}X'\Omega^{1/2})\Omega^{-1/2}X$$ Now let $$C:=\Omega^{-1/2}X$$ and $$D:=\Omega^{1/2}X$$ so that we may write $$C'(I-D(D'D)^{-1}D')C$$ Now, $$M_D:=I-D(D'D)^{-1}D'$$ is a residual-maker, hence, symmetric and idempotent projection matrix. Let $$c:=Ce$$ for some nonzero vector $$e$$. Then, for $$f:=M_Dc$$, $$c'M_Dc=c'M_D'M_Dc\equiv f'f\geq0,$$ as $$f'f=\sum_if_i^2$$ is a sum of squares.
Now, if $$X$$ actually were square (and invertible), we would have $$X'X(X'\Omega X)^{-1}X'X=X'XX^{-1}\Omega^{-1}(X')^{-1}X'X=X'\Omega^{-1} X$$ so that OLS and GLS would be equally efficient!
In fact, they would be numerically identical, as $$(X'\Omega^{-1} X)^{-1}X'\Omega^{-1}y=X^{-1}\Omega X'^{-1}X'\Omega^{-1}y=X^{-1}y=X^{-1}X'^{-1}X'y=(X'X)^{-1}X'y$$