I have an expression of the form:
$ACA^{'}$
where $C$ is an invertible, symmetric and positive definite matrix. I'm trying to figure out if the expression above is invertible. The $C$ matrix is $n \times n$ and the $A$ matrix is $k \times n$. Every row of A has exactly one entry =1 and one entry =-1, all other entries are 0.
Any help, especially something pointing to a theorem, would be greatly appreciated.
If $C\in\mathbb{R}^{n\times n}$ is (symmetric and) positive definite and $A\in\mathbb{R}^{k\times n}$ with $k\leq n$, then $ACA^T\in\mathbb{R}^{k\times k}$ is invertible if and only if $A$ has full rank. (You can think of $ACA^T$ as the projection of $C$ onto the subspace spanned by the rows of $A$, so it makes sense to expect them to be linearly independent.)
To see this, let $x\in\mathbb{R}^k\setminus\{0\}$ be arbitrary. If $A$ has full (row) rank, then $A^T$ has full (column) rank as well and thus is injective. Hence, $y:=A^Tx \in\mathbb{R}^n\setminus\{0\}$, and the positive definiteness of $C$ yields $$ x^T(ACA^T)x = (A^Tx)^TC(A^Tx) = y^TCy > 0, $$ i.e., $ACA^T$ is (symmetric and) positive definite and thus invertible. Conversely, if $A$ does not have full rank, $A^T$ is not injective and there exists a vector $x\in\mathbb{R}^k\setminus\{0\}$ with $A^Tx = 0$ and hence $ACA^Tx = 0$. It follows that $ACA^T$ is not injective and thus not invertible.
