= Solution
Both $L$ and $R$ have $m$ <linearly independent> columns. The <Gram matrix> $R^\top R$ is invertible because
$$
y^\top R^\top Ry=\|Ry\|^2>0\qquad(y\ne0).
$$
Consequently $R^\top$ is a <surjective linear map>: for any $y\in\mathbb R^m$, the choice $x=R(R^\top R)^{-1}y$ satisfies $R^\top x=y$. Hence
$$
\operatorname{im}B=L(\operatorname{im}R^\top)=\operatorname{im}L.
$$
The latter <image of a linear map> has <dimension> $m$, giving \b[$\operatorname{rank}B=m$]. If $m=0$, $B=0$ and the conclusion is immediate. This argument explains why the product retains its full possible <matrix rank>; a general product of two <matrices> need not do so.
Back to article page
