Inverse of projected matrix in terms of original inverse matrix

Question

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 $C$?

Answer 1

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 $$ (C^T M C)(T^{-1} S M^{-1} S^T (T^{-1})^T) = C^T S^T (T^{-1})^T = T^T (T^{-1})^T = I$$