Inverse of projected matrix in terms of original inverse matrix Let $M$ be an invertible complex $n\times n$ matrix and let $C$ be a complex $n\times (n-1)$ matrix.  Write $\tilde{M} = C^T M C$.  Under what conditions is $\tilde{M}$ invertible and, if those conditions hold, how can $\tilde{M}^{-1}$ be written in terms of $M^{-1}$ and $C$?
 A: Since the rank of a product is at most the rank of one of the factors, $C$ must certainly have rank $n-1$.  Now if $C$ has rank $n-1$, its null space is $\{0\}$, while the null space of $C^T$ is one-dimensional.  Let $u$ be
a nonzero vector in that null space, and let $v = M^{-1} u$.  Then the requirement for $C^T M C$ to be invertible is that $v$ is not in the range of $C$, i.e. that the
matrix obtained by adjoining $v$ as a column to $C$ is invertible.
EDIT:
Suppose this condition is satisfied.  The projection of $\mathbb C^n$ on the range of $C$ that annihilates $v$ is
$$P = I - \dfrac{v u^T}{u^T v}$$
Note that since $u^T w = 0$ for all $w \in \text{Ran}(C)$, and $v$ and $\text{Ran}(C)$ span all of $\mathbb C^n$, if $u^T v = 0$ we'd have $u = 0$, which is not the case; therefore $u^T v \ne 0$.
Now $C$ has rank $n-1$, so some $n-1$ rows of $C$ are linearly independent; for convenience, let's say these are the top $n-1$ rows.
Let $T$ be the $(n-1) \times (n-1)$ matrix formed by these
rows.  Let $S$ be the $(n-1) \times n$ matrix formed by the corresponding rows of $P$.  This satisfies $S C = T$ while $S v = 0$. 
Now I claim that
$$(C^T M C)^{-1} = T^{-1} S M^{-1} S^T (T^{-1})^T$$
To verify this, note that 
$C T^{-1} S v = 0$, while for $y = C x \in \text{Ran}(C)$ we have
$$C T^{-1} S y = C T^{-1} S C x = C T^{-1} T x = C x = y$$
Thus $C T^{-1} S = P$, and 
$$ (C^T M C)(T^{-1} S M^{-1} S^T (T^{-1})^T) = C^T M P M^{-1} S^T (T^{-1})^T$$
Now 
$$M P M^{-1} = I - \dfrac{M v u^T M^{-1}}{u^T v} = I - \dfrac{u u^T M^{-1}}{u^T v}$$
but $C^T u = 0$ so
$C^T M P M^{-1} = C^T$.  Thus we get
$$ (C^T M C)(T^{-1} S M^{-1} S^T (T^{-1})^T) = C^T S^T (T^{-1})^T = 
T^T (T^{-1})^T = I$$
