Context:
Consider matrices $A \in \mathbb{R}^{m \times n}$ and $B \in \mathbb{R}^{m \times m}$ where $B$ is invertible. Let where $A$ have $k$ zero columns, so $A$ has the following form:
$$A = \begin{bmatrix} A_{00} & 0_{m \times k} \end{bmatrix}$$
where $A_{00} \in \mathbb{R}^{m \times n-k}$. Now, consider $A^T B A$, which is given by $$ \begin{align} A^T B A = \begin{bmatrix} A_{00}^T B A_{00} & 0_{n-k \times m} \\ 0_{k \times n-k} & 0_{k \times k} \end{bmatrix} \end{align} $$ where $A_{00}^T B A_{00} \in \mathbb{R}^{n-k \times n-k}$.
So, given the fact that $A$ has at least one zero column, then $\det[A^T B A]=0$.
Question:
Now, consider $A \in \mathbb{R}^{m \times n}$ (where $A$ does not necessarily have zero columns) and $B \in \mathbb{R}^{m \times m}$ where $B$ is invertible.
If $\det[A^T B A]=0$, does a similarity transform $T$ exist such that $TA$ has the previously mentioned form? For example, does $T$ exist such that $AT = \begin{bmatrix} A_{00} & 0_{m \times k} \end{bmatrix}$.
This is related to a problem I am working on in estimation theory (more specifically related Kalman decomposition), but I do not think the details of that overall problem is necessary for this question. I feel this result may be obvious for someone more experienced in linear algebra, but I am stuck.