"Independence" with respect to a particular linear combination? Suppose $A$ is an $m \times m$ invertible square matrix. $B$ is an $m \times n$ matrix $(m<n)$ with rows $R_1,...,R_m$ which satisfy the following property $P$: 
There exist no set of scalars $\lambda_1,...,\lambda_m$, such that $ \lambda_i \in \{-1,0,1\} \; \forall i = 1(1)m, \lambda_i \neq 0$ for some $i$, such that $\lambda_1R_1 + ... + \lambda_mR_m = 0$. 
Simply put, no row can be expressed as a combination of sums and differences of any subset of the other rows. 
$A$ obviously satisfies $P$ since its rows are linearly independent. 
My question is, can we say anything about whether $AB$ satisfies $P$ or not? (Something analogous to the property that if $A$ and $B$ have linearly independent rows and are of compatible dimensions then $AB$ has linearly independent rows.)
 A: We can relate the linear independence of row/column vectors to invertible property of matrix. 
It's easier if we work with column vectors rather than row vectors. 

Proposition 1. If $m \times n$ matrix $\mathbf{B}$ that satisfies property $P$ then so does $\mathbf{AB}$ where $A$ is an $m \times m$ invertible matrix.

Let column vectors in $m \times n$ matrix $\mathbf{B}$ be $\mathbf{b}_i \; (1 \le i \le n)$. Since $\mathbf{B}$ satisfies $P$ so there doesn't exists non-all-zero $x_1, \ldots, x_n \in \{-1,0,1\}$ so $\displaystyle \sum_{i=1}^n x_i \mathbf{b}_i=0$. On the other hand, if we let $\mathbf{x}=(x_i)_{n \times 1}$ then we have $\mathbf{Bx}=\sum_{i=1}^n x_i\mathbf{b}_i$. This follows that there doesn't exists a solution $\mathbf{x}$ for $\mathbf{Bx}=0$ so entries of $\mathbf{x}$ are from $\{-1,0,1\}$ and $\mathbf{x} \ne 0$. 
With similar argument, it suffices to prove that there doesn't exist such solution $\mathbf{x}$ to $\mathbf{ABx}=0$. Note that $\mathbf{ABx}=\mathbf{A(Bx)}$ and since $\mathbf{A}$ is invertible, which means only solution for $\mathbf{Ay}=0$ is $\mathbf{y}=0$, we find $\mathbf{Bx}=0$. Combining with the above assumption, we are done. 

Change it back to row vectors: If $\mathbf{B}$ satisfies $P$ then so does $\mathbf{BA}$.

We claim that $\mathbf{BA}$ satisfies $P$ if $\mathbf{B}$ satisfies $P$ with $A$ is $n \times n$ invertible matrix.
This time, consider column vectors of $B^T$ and $A^TB^T$ instead. If all row vectors of $B$ satisfy property $P$ then all column vectors of $B^T$ satisfy property $P$, which means all column vectors of $A^TB^T$ satisfy property $P$. And since $A^TB^T=(BA)^T$, we find row vectors in $BA$ satisfies property $P$.

What about row vectors in $\mathbf{AB}$? Does $\mathbf{AB}$ satisfy $P$ if $\mathbf{B}$ does?

This case is different from $\mathbf{BA}$ because we can't apply proposition 1 for this, as $(AB)^T=B^TA^T$, i.e. the invertible matrix is at the right not at the left like in proposition 1. I need to put more thought into this part...
