Is $\mathbf {B^TAB}$ non-singular for a non-singular $\mathbf A$, and $\mathbf {B}$ with full column-rank? If $\mathbf A$ is any square non-singular matrix of dimension $n \times n$. And $\mathbf B$ is a $n \times m$ matrix with $\mathrm{rank(\mathbf B)} = m$. Is the full rank condition of matrix $\mathbf B$ both sufficient as well as necessary to state that the matrix product $\mathbf {B^TAB}$ is non-singular?
i.e., can we write:
$ \mathbf {A}$ is non-singular $\iff$ $\mathrm{rank(\mathbf B)} = m$ and $\mathbf {B^TAB}$ is non-singular
 A: Nope, the statement is not true in this form. 
A simple example is:
$$
B=\begin{bmatrix}1\\1\end{bmatrix},\quad
A=\begin{bmatrix}1&0\\0&0\end{bmatrix}.
$$
Then $\mathrm{rank}(B)=1$ and $B^TAB=1$ is non-singular, but $A$ is singular.
If
$$
B=\begin{bmatrix}1\\1\end{bmatrix}, \quad
A=\begin{bmatrix}1&0\\0&-1\end{bmatrix},
$$
then $\mathrm{rank}(B)=1$, $A$ is non-singular, but $B^TAB=0$ is singular.
The condition $B^TAB$ and $\mathrm{rank}B=m$ does not need to be sufficient for the non-singularity of $A$ since the conditions impose a sort of "non-singularity on a restricted subspace", which obviously does not need to be sufficient for the "non-singularity of the whole". On the other hand, a non-singular $A$ does not imply that $x^TAx\neq 0$ (e.g., indefinite symmetric matrices). 
NOTE: $\mathrm{rank}(B)=m$ should actually be an assumption for the equivalence (if there was any) as the non-singularity of $A$ and the full rank of $B$ are completely independent things (non-singularity of $A$ does not imply anything about the rank of $B$).
