For which $B$ is $BA$ positive definite? Suppose $A$ is an $m\times n$ real-valued full-rank matrix. What is a nice way to characterize the set of real-valued $n\times m$ matrices $B$ such that $BA$ is positive definite?
For $n=1$, this is the set of vectors in the same half-plane as $A$ (viewed as column vector).

Here are two characterisations of all matrices $B$ that make $BA$ positive definite.

*

*A more parsimonious characterisation: $B=SA^++Z^T$ for some positive definite matrix $S$ and some $m\times n$ matrix $Z$ such that $Z^TA=0$.

*A nicer-looking characterisation: $B=A^TP$ for some positive definite matrix $P$.

Pick any matrix $\widehat{A}$ with $m-n$ columns such that its column space is the orthogonal complement of the column space of $A$. Then $A^TA\in\mathbb R^{n\times n}$ and $\pmatrix{A&\widehat{A}}\in\mathbb R^{m\times m}$ are invertible. Therefore we may write
$$
B=A^TA\pmatrix{X&Y}\pmatrix{A^T\\ \widehat{A}^T}
=A^TA\left(XA^T+Y\widehat{A}^T\right)
$$
for some square matrix $X$ and $n\times(m-n)$ matrix $Y$. Hence $BA=A^TAXA^TA$ is positive definite if and only if $X=(A^TA)^{-1}S(A^TA)^{-1}$ for some positive definite matrix $S$, and if this is the case, then
$$
B=A^TA\left(XA^T+Y\widehat{A}^T\right)
=SA^++Z^T
$$
where $Z^T=(A^TA)^{-1}Y\widehat{A}^T$ is a matrix such that $Z^TA=0$. Furthermore, as $A^T\widehat{A}=0$,
\begin{aligned}
B=A^TA\left(XA^T+Y\widehat{A}^T\right)
&=A^T\,\underbrace{\pmatrix{A&\widehat{A}}\pmatrix{X&Y\\ Y^T&kI}\pmatrix{A&\widehat{A}}^T}_P.
\end{aligned}
So, by picking a large $k>0$, we may also write $B=A^TP$ for some positive definite $P$.
Conversely, if $B=SA^++Z^T$ for some positive definite $S$ and some matrix $Z$ such that $Z^TA=0$, then $BA=S$ is positive definite; if $B=A^TP$ with a positive definite $P$, then $BA=A^TPA$ is also positive definite.
