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).
 A: In general you need $m \ge n$ as otherwise the rank of $BA$ is at most $m < n$. For me positive definite means symmetric and all positive eigenvalues.
This is not a full characterization.
One general class of matrices is as follows. Let
$$ \hat{\Sigma} = \begin{bmatrix} D & C \end{bmatrix}$$
be $n \times m$ matrix where $D$ is a $n \times n$ diagonal with positive diagonal entries and $C$ is any $n \times (m-n)$ matrix. Let $A = U\Sigma V^T$ be the SVD, then $B = V\hat{\Sigma}U^T$ results in $BA = V\hat{\Sigma}\Sigma V^T$ is positive definite.
Hence $B$ lives in at least $n(m-n)$ dimensional linear space (for $C$) product with $\mathbb{R}_+^n$ (for $D$).
We have the whole above space as for two $D_1,D_2$ and $C_1, C_2$ such that $\begin{bmatrix} D_1-D_2 & C_1-C_2 \end{bmatrix} \neq 0$
$$\left\|V\begin{bmatrix} D_1 & C_1 \end{bmatrix}U^T -V\begin{bmatrix} D_2 & C_2 \end{bmatrix}U^T \right\|_F^2 = \left\|\begin{bmatrix} D_1-D_2 & C_1-C_2 \end{bmatrix} \right\|_F^2 > 0 .$$
Where that first equality is due unitary invariance of Frobenius norm.
A: 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.
