Is a vector a multiple of some other vector? Let $\bf a_1,\ldots, a_m$ be vectors in $\mathbb{R}^n$ where $n \geq 2$. Does there exist 
$$
B=\left[\begin{array}{cccc} 1 & 0 & \cdots & 0 \\ 
                            0 & b_1 & \cdots & b_{n-1} \end{array}\right]
$$
such that $B {\bf a_1}=\left[\begin{array}{c} a_{11} \\ 
                            \sum_{j=2}^n b_ja_{1j} \end{array}\right]$ is not a multiple of $B {\bf a_i}=\left[\begin{array}{c} a_{i1} \\ 
                            \sum_{j=2}^n b_ja_{ij} \end{array}\right]$ for all $i=2,\ldots, m$?
I think the answer is that such a $B$ exists if and only if $\bf a_1$ is not a multiple of $\bf a_i$ for all $i=2,\ldots, m$. I'm sure of sufficiency (the proof seems obvious but can't quite put arguments together) but not sure of necessity.
P.S. Not sure if the "Title" of this question is a good one.
Edit: The question is answered in @Omnomnomnom notation replicated here.
We have a matrix
$$
A = \pmatrix{a_{11} & \cdots &a_{1n}\\ \mathbf{v}_1 & \cdots & \mathbf v_{n}}
$$
with $\mathbf v_i \in \Bbb R^{n-1}$, and we would like to select
$$
B = \pmatrix{1 & 0\\0& \mathbf b ^T}
$$
where $\mathbf b \in \Bbb R^{n-1}$ such that the first column of $BA$ is not a multiple of any other.

In fact, if we can suppose $a_{1i} \neq 0$, it is equivalent to ask the same question about the product $B\tilde A$, where we define
$$
\tilde A = \pmatrix{1 & \cdots & 1\\ \frac{1}{a_{11}}\mathbf{v}_1 & \cdots & \frac{1}{a_{1n}}\mathbf v_{n}} := 
\pmatrix{1 & \cdots & 1\\ \tilde v_1 & \cdots & \tilde v_n}
$$
In this case, two columns will only be multiples if they are also equal.
 A: Solution in the case of $a_{1i} \neq 0$ for all $i$:
Suppose that $a_1$ is a multiple of $a_i$ for some $a_i$.  This means that $\tilde v_1 = \tilde v_i$.  It follows that $b^T \tilde v_1 = b^T \tilde v_i$ for any choice of $b$.
Now, suppose that no suitable $B$ exists. That is, suppose that our $\tilde v_i$ are such that for any $b$, there is a choice of $i > 1$ such that $b^T \tilde v_1 = b^T \tilde v_i$. That is, suppose that there is a set $S = \{\tilde v_{i_1},\dots,\tilde v_{i_k}\}$ such that $b^T \tilde v_1 = b^T \tilde v_{i_k}$ for some choice of $k$, where the choice of $k$ may depend on $b$.
For any $v \in S$, we define K_v the set of vectors $b$ such that $b^T v = b^T \tilde v_1$.  That is, $K_v$ is the set of vectors $b$ such that $b^T(v - \tilde v_1) = 0$.  That is, K_v is the kernel of the linear transformation $T(x)$ defined by
$$
T(x) = x^T(v - \tilde v_1)
$$
which, in particular, means that each $K_v$ is a subspace of $\Bbb R^{n-1}$. 
Now, define $K = \cup_{v \in S} K_v$.  The above statement is equivalent to the statement $K = \Bbb R^{n-1}$.  However, the union of finitely many proper subspaces of $\Bbb R^{n-1}$ cannot be equal to $\Bbb R^{n-1}$ (remember: the union of two subspaces is only a subspace if one contains the other).  Thus, there exists a $v \in S$ such that $K_v = \Bbb R^{n-1}$.  
That is, there is an $i>1$ such that $b^T v_1 = b^T v_i$ for every $b \in \Bbb R^{n-1}$.  By considering $b = (1,0,\dots,0)^T, b = (0,1,0,\dots,0)^T,$ and so forth, we conclude that $v_1 = v_i$.
Thus, $a_1$ must be a multiple of some $a_i$ for $i>1$, as desired.

As for the other cases: note that, assuming $b \neq 0$ and $a_1 \neq 0$, $B a_1$ can only be a multiple of $B a_i$ if the first entries of $a_1$ and $a_i$ are both either zero or non-zero.  Using this fact, we may conclude that the necessity and sufficiency of your condition holds in general.
