Consider a real, symmatric and positive definite $n \times n$ matrix $\mathbf{A}$, and a $n \times m$ matrix $\mathbf{W}$. $\mathbf{W}$ contains $m$ columns with all zeros except a single entry in each column which is $1$.
If $\mathbf{A}$ is positive definite then $\mathbf{A}^{-1}$ is positive definite, but is it possible to say anything general about the definiteness of the matrix
$$\mathbf{W}^T \mathbf{A}^{-1} \mathbf{W}$$
Example: Consider
$$\mathbf{A}^{-1} = \begin{bmatrix} a_{11} & a_{12} & a_{13} \\ a_{21} & a_{22} & a_{23} \\ a_{31} & a_{32} & a_{33} \end{bmatrix}$$
and
$$\mathbf{W} = \begin{bmatrix} 0 & 1 \\ 0 & 0 \\ 1 & 0 \end{bmatrix}$$
then
$$\mathbf{A}_w = \mathbf{W}^T \mathbf{A}^{-1} \mathbf{W} = \begin{bmatrix} a_{33} & a_{31} \\ a_{13} & a_{11} \end{bmatrix}$$
Is it possible to say anything about the definiteness of $\mathbf{A}_w$?
In this case $A_w$ is positive definite. If $B$ is positive definite then $W^TBW$ is always positive semidefinite, and is positive definite if the columns of $W$ are linearly independent.
In general, for a column vector $x$, $$x^T(W^TBW)x=(Wx)^TB(Wx)\ge0$$ as $B$ is positive definite. If in addition, $x\ne 0$ and $W$ has linearly independent columns, then $Wx\ne0$ and $$x^T(W^TBW)x=(Wx)^TB(Wx)>0.$$