in order to solve a little problem that give me a friend, I have the following question: Is it possible to find a subspace $W$ of the $\mathbf{Z}/2\mathbf{Z}$-vector space $(\mathbf{Z}/2\mathbf{Z})^{64}$ such that:
$\bullet \ e_1$ is in $W$
$\bullet \ \mathrm{dim}(W)=58$
$\bullet$ all the elements with exactly two nonzero coefficients are outside of $W$.
Thanks a lot!
For the ones who are interested in the problem: You and a friend are prisoners. To get free, the warders play a game with you: You know that you are going to go the next day in another room, with a chess board and coins which show head or tail randomly on every square. Your kidnapper shows you under which coin the key of your prison is hidden. You have to turn over exactly one coin after you are leaving the room. After that, your friend comes in the room and has to find the key in one try. Which strategy can be used?
Indeed, such a subspace exists. Let $f:\{0,\dots,63\} \to \{0,1\}^6$ denote the map for which $$ f(k) = (f_0(k),\dots,f_5(k)) \iff k = \sum_{i=0}^5 f_i(k) 2^i. $$ Let $W$ be the kernel of the map $\phi:\Bbb Z_2^{64} \to \Bbb Z_2^6$ defined by $$ \phi(e_{k}) = f(k-1), \quad k = 1,2,\dots,64. $$ We see that no two elements lie in the same equivalence class because $f(e_j) \neq f(e_k)$ whenever $j \neq k$.