$m$ binary vectors in $\mathbb{R}^n$ with $v_i.v_j=k$ are linearly independent 
Let $m<n$ and $v_1,\cdots,v_m$ be distinct binary vectors in $\mathbb{R}^n$ and
  $k\in\mathbb{N}$. If for each $i\neq j$, we have $v_i.v_j=k$, prove
  that these vectors are linearly independent.

If we define $N_j=\text{Number of 1s in $v_j$}$ and suppose $v_1=\sum_{i\geq2}\lambda_iv_i$, we'll have :
$$\left\{\begin{array}{l}N_1=k\sum_{i\geq2}\lambda_i\\
\displaystyle\forall j\geq2:\;k=N_j+k\sum_{i\neq j}\lambda_i
\end{array}\right.$$
I think some inequalities might be applied for contradiction.
I don't think induction may help us a lot, since I tried for several minutes.
 A: You definitely have the right idea! but it will be easier to see the proof through if you keep the equations symmetric. In other words, instead of writing one vector as a linear combination of the others, just write down a linear combination $\sum_{j=1}^m \lambda_j v_j$ that equals the zero vector (we need to show that all the $\lambda_j$ equal $0$). Now taking dot products with each $v_i$ in turn yields the $m$ equations $k\lambda_1 + \cdots + k\lambda_{i-1} + N_i + k\lambda_{i+1} + \cdots + k\lambda_m = 0$.
Solving this system of equations for the $\lambda_j$ is equivalent to finding the null space of the $m\times m$ matrix whose diagonal entries are $N_1,\dots,N_m$ and whose off-diagonal entries all equal $k$. An obvious sequence of row reductions leads to the equivalent matrix
$$
\begin{pmatrix}
N_1 & k & k & \cdots & k \\
-(N_1-k) & N_2-k & 0 & \cdots & 0 \\
-(N_1-k) & 0 & N_3-k & \cdots & 0 \\
\vdots & \vdots & \vdots & \ddots & \vdots \\
-(N_1-k) & 0 & 0 & \cdots & N_m-k
\end{pmatrix},
$$
and another obvious sequence of row reductions leads to a lower triangular matrix with $N_j-k$ on the diagonal ($2\le j\le m$) and a complicated expression in the upper-left corner.
Now, use the key fact that the assumptions imply that $N_j > k$ for all $j$ (obvious after a little thought). Therefore $N_j-k$ is positive for all $k$, and (after writing down the complicated upper-left entry) this implies that all the diagonal entries of this matrix are positive. Therefore its null space is trivial: the only solution is $\lambda_1=\cdots=\lambda_m=0$, as desired.
(And yes, we didn't have to explicitly assume $m\le n$ - it came out of the proof.)
