Is $Q^{-1} - E(E^TQE)^{-1}E^T$ always positive definite? If $Q \in \mathbb{R}^{m \times m}$ is symmetric and positive definite and $E \in \mathbb{R}^{m \times n}$ with $m > n$ has rank $n$, then is 
$Q^{-1} - E(E^TQE)^{-1}E^T$  
always positive definite? It appears to be true for a number of sample instances that I have tested, but I can't see how how to prove it in general. 
 A: No. $Q$ is the 2 by 2 identity matrix, $E = (1,0)^T.$
A: Write $Q = U^T V U$ with orthogonal $U$ and diagonal positive $V$.
Then $Q^{-1} - E ( E^T Q E )^{-1} E^T = U^T V^{-1} U - E ( (U E)^T V (U E) )^{-1} E^T = U^T ( V^{-1} - U E ( (UE)^T V (UE) )^{-1} (UE)^T  ) U$.
In other words, you can assume, without loss of generality, $Q$ to be diagonal.
So assume $Q$ is diagonal. Then $Q = S^2$ for a positive $S$. By means of a similar trick, you can assume that $Q$ is the identity.
Now $P := I - E (E^T E)^{-1} E^T$ is positive semi-definite if $E$ is injective: the second part is the orthogonal projection onto the subspace spanned by the columns of $E$, so $P$ is an orthogonal projection to its complement. 
In particular, $P$ need not be positive definite, and in your case it is not.
The original expression was probably an orthogonal projection with respect to the scalar product induced by $Q$.
A: Your matrix is equal to
$Q^{-1/2} \left[ I - Q^{1/2}E \left( E^TQ^{1/2} Q^{1/2}E \right)^{-1}  E^TQ^{1/2} \right] Q^{-1/2}$. So, if we put $F=Q^{1/2}E$, we may consider $I_m - F (F^T F)^{-1} F^T$ instead, where $F$ has full-rank. Perform singular value decomposition on $F$, we may further assume that $F$ is a rectangular diagonal matrix. Hence $I_m - F (F^T F)^{-1} F^T=\pmatrix{0_{n\times n}\\ &I_{(m-n)\times(m-n)}}$, which is always positive semidefinite but never positive definite.
