Is $BAB'$ (with positive definite $A$ and full-row rank$(B) = k$) itself of rank $k$? Here's the setup:
A matrix $B$ with dimensions $k \times p$ with $k \leq p$ and rank$(B) = k$.
A matrix $A$ with dimensions $p \times p$ is positive definite (not necessarily symmetric).
Question: Is the square matrix $BAB'$ (which will have dimensions $k \times k$) a full-rank matrix?
I know that the rank is at most $k$, but if the rank can be shown to equal $k$, that will make things very easy for what I need to do.
Note: $B'$ is the notation I use for the transpose of matrix $B$, partially because I suck at HTML.
Thank you in advance.

Hint: motivate the following steps to prove your statement
$BAB^Tx=0$ $\Rightarrow$ $x^TBAB^Tx=0$ $\Rightarrow$ $y^TAy=0$ $\Rightarrow$ $y=B^Tx=0$ $\Rightarrow$ $x=0$.
