Proof singularity of non square matrix times square matrix times transpose of the non square matrix: $\det(ABA^T)=0$

Question

Can somebody help me with the proof of $\det(ABA^t)=0$, in case of a non-square matrix $A$ with dimension $(n \times m), n>m$ and a square matrix $B$ of dimension $(m \times m)$. Thank you for your help!

Answer 1

The rank of $B$ is at most $m$. Matrix multiplication cannot increase the rank, so $ABA^t$ has rank at most $m$. Since $m<n$ it follows that the $n\times n$ matrix $ABA^t$ has rank less than $n$, so its determinant is $0$.