TITLE: My fun conjecture about linearly independence
QUESTION [6 upvotes]: In the $\mathbb{R}^n$ vector space, there are distinct $m$ vectors $v_i$'s ($1< i\leq m)$
such that each component has value 0 or 1.
Let $A_i$ be the set of $j$'s where $j$-th component of $v_i$ is 1.
Also, for each $i \neq j$, $A_i$ and $A_j$ has common $k$ elements. Where $k$ is a given integer $1\leq k <n$.
For example, when $n=3, k=1$. $v_1=(1,1,0), v_2=(1,0,1), v_3=(0,1,1)$ satisfy those conditions since $A_1=\{1,2\},A_2=\{1,3\},A_3=\{2,3\}$.
My conjecture is : those $v_i$'s are linearly independent.
With some rough programming, this conjecture was true when $n \leq 10$. 
I tried to prove this conjecture with induction on $k$, but I failed.
*Some people misunderstood question. 
Actually question is : For given $n,m,k$, is every families of vectors with above condition are linearly independent.
Can you prove or disprove this conjectrue?

REPLY [5 votes]: Your condition says that $\langle v_i,v_j\rangle = k$ when $i\ne j$, and let's write $\langle v_i,v_i\rangle = n_i$. Let $A$ be the matrix having the $v_i$ as columns. Then 
$$A^TA = \begin{pmatrix}n_1 & k & k & \cdots & k \\
                          k & n_2 & k & \cdots & k \\
                            &  & \ddots & & \\
                          k & k & k & \cdots & n_m
\end{pmatrix}$$
Note that at most one of the $n_i$ can be equal to $k$ because any two of them would have to be equal. Also note that we have to explicitly exclude the case that $v_i = 0$ for some $i$. Otherwise we are done: $A^TA$ is clearly invertible so $A$ can not have a non-trivial kernel, i.e. no non-trivial linear combination of its columns, the $v_i$, can give $0$.