What can be said about invertibility of $ABA^T$ if $A$ has full rank and $B$ is invertible? I have an $n\times k$ matrix $A$ of rank $n$ with $k>n$ and an invertible $k\times k$ matrix $B$ which is unfortunately not symmetric or anything special like that. What can I say about the rank of $ABA^T$ under these circumstances, if anything? Are there reasonable conditions besides $B$ positive definite symmetric which will make this product invertible?
In my set up, $A,B$ are natural number matrices with the main diagonal of both $B$ and $ABA^T$ having positive entries if that matters.  Moreover, $A$ is surjective as a homomorphism between $\mathbb Z^k$ and $\mathbb Z^n$.  I'm even willing to assume that $B$ has determinant $\pm 1$, which it does in many but not all cases of interest.
 A: You cannot say anything in the general case. For instance try
$$
A=\begin{bmatrix} 1&0\end{bmatrix},\qquad\qquad B=\begin{bmatrix} 0&1\\ 1&0\end{bmatrix}.
$$
Then $B$ is invertible and symmetric, $A$ has full rank, and
$$
ABA^T=\begin{bmatrix} 1&0\end{bmatrix}\begin{bmatrix} 0&1\\ 1&0\end{bmatrix}\begin{bmatrix} 1\\0\end{bmatrix}=0.
$$
This particular game can be played whenever $B$ has a zero in its diagonal in some basis (equivalently, if the numerical range of $B$ contains zero).
A: For vector space homomorphisms $A: V \rightarrow W$ we have the following identity:
\begin{equation}
dim(V) = rg(A) + dim(ker(A))
\end{equation}
In your case we have: $V = \mathbb{Z}^k, W = \mathbb{Z}^n$ and $rg(A)=n$, with $dim(V) = k$, $dim(W) = n$, $ rg(A^T) = n$. Moreover $A^T: W \rightarrow V$.
The strategy is to try to find more information about the kernel of $ABA^T$. Using the identity above for the transposed matrix we arrive at:
\begin{equation}
dim(W) = n = n + dim(ker(A^T)) \Rightarrow dim(ker(A^T)) = 0
\end{equation}
So the only vector contributed to the kernel by $A^T$ is $\{0\}$. Next let's look at $B$. $B$ is an isomorphism, so again the only element of its kernel is $\{0\}$. So far we only have $ker(ABA^T) = \{0\}$. The only thing left to do is to check the kernel of $A$. The following holds for the latter:
$k = n + dim(ker(A)) \iff dim(ker(A)) = k-n$
Because $A^T$ is not surjective, we have to ask ourselves whether $ker(A) \subseteq Image(A^T)$. This is only the case when
$dim(Image(A^T)) = n \geq k-n = dim(ker(A))$
so $2n \geq k$. This means that when $k > 2n$, $ker(ABA^T) = \{0\}$ and so $rg(ABA^T) = n$. In other words, the matrix $ABA^T$ is invertible when $k>2n$.
I couldn't generalize any more for the case in which $2n>k$. If you have any suggestions, let me know and I'll edit my answer.
A: If you also have that $\text{Col}(BA^T)=\text{Col}(A^T),$ then $ABA^T$ is invertible.
Why?
Well, assume $ABA^Tx=0$. Then $BA^Tx\in \mathcal{N}(A)=\big(\text{Col}(A^T)\big)^{\perp}=\big(\text{Col}(BA^T)\big)^{\perp}$. But $BA^Tx$ clearly belongs to $\text{Col}(BA^T)$ so $BA^Tx$ must equal zero. Since $B$ is invertible, $A^Tx=0$, and since every column of $A^T$ is independent, $x=0$.
