Linear algebra and combinatorics. For a family with even size sets and even intersections prove that $|F| \le 2^{n/2}$ Let $F \subset P(n)$ be a family such that for all i and j
$ |f_i \cap f_j|$ and $|f_i|$ are even
Prove that $|F| \le 2^{n/2}$
Now I think we go by contradiction and say if $|F| \ge 2^{n/2}+1$ then, using characteristic functions as in the proof of Fisher's inequality, we must have at least $n/2 + 1$ linearly independent vectors. Now what?
Any help appreciated!
 A: I will show you another approach. First we will work on $\mathbb{F}_2^n$, and define $v_i$ to be the characteristic vector of $f_i$.
Consider $V$ the span of the $v_i$, and notice that because we have $v_i \cdot v_j = 0 $, and $(v_i + v_j) \cdot v_k = 0$ and every vector of $V$ fall into one of these cases then $V \subseteq V^\perp$.
But we have that $ n = \dim(V) + \dim(V^\perp) \geq 2\dim(V) $, so $\dim(V) \leq n / 2$. 
We conclude from here that $V$ has at most $2^{\lfloor n/2 \rfloor}$ vectors (you have two choices for each entry), and the $v_i$ are contained in these, so the inequality follows.
Btw, this seems like a big bound, especially because if you switch the restriction that the $|f_i|$ are even to odd, then your new bound will become $n$, much less. But this is in fact sharp as you can first divide the elements of $\{ 1,2, \ldots , n\}$ into pairs, and then take your family to be all possible unions of these, then you end up with $2^{n/2}$ elements in the family, and they satisfy the conditions.
A: First, consider $\mathbb{F}_2^2$. Then by the answer of alejopelaez, we should have $V=span_\mathbb{R} \{(1,1)\}$ and ${V}^\perp = span_\mathbb{R} \{(1,0),(0,1)\}$. See, the formula $n = \dim(V) + \dim(V^\perp)$ does not hold. Use Fisher's inequality which is given by: Let $S=\{1,2,.,.,.,n\}$ and $\mathcal{F}\subset P(S)$ (power set of $S$) and L be a set of natural numbers such that $\forall A,B \in \mathcal{F} $, $|A\cap B| \in L$. Then
\begin{equation*}
|\mathcal{F}| \leq \sum_{0\leq i\leq|L|}{^nC_i}.
 \end{equation*}
And take $L=\{0,2,4,6,.,.,.,\}$. So $|L|=[\frac{n}{2}]$, where   $[x]$ is the greatest integer $\leq x$.  
