Do $J$, the all-ones matrix of even order, always have eigenvectors consisting of entries $-1, 1$ only? 
Do  $J$, the all-ones matrix of even order, always have eigenvectors consisting of entries $-1, 1$ only?

It seems so, vector having all its entries $1$ is one eigenvector for larest eigenvalue $n$ and any vector having half its entries as $1$ and half as $-1$ is eigenvector for ev $0$.
For $n=4k$ we have Hadamard matrices there, so those vector will work.
But I don't know how to show existence of such $n$ linearly independent vectors.
 A: Yes. Let $S$ be the subset of $\{-1, 1\}^{2n}$ where each vector $v$ in $S$ has $n$ $1$s and $n$ $-1$s. It suffices to show that $\mathrm{span}(S)$ is equal to the $2n-1$ dimensional subspace perpendicular to the vector $(1,1,\dots, 1)$. (As you've noted, the all one's vector $(1,1,\dots, 1)$ already spans the other one-dimensional eigenspace of $J$). 
To see this, we'll show that for each $i\neq j$, the element $e_i - e_j$ belongs to $\mathrm{span}(S)$ (where $e_i$ is the $i$th unit vector). Let $v$ be an element of $S$ that has a $1$ in position $i$ and a $-1$ in position $j$, and let $v'$ be the vector obtained from $v$ by swapping the elements in position $i$ and $j$. Note that $v'$ is also in $S$ and $(v-v')/2 = e_i - e_j$; hence $e_i - e_j \in \mathrm{span(S)}$.
Now, if $v = (v_1, v_2, \dots, v_{2n})$ is orthogonal to $(1,1,\dots,1)$, then $\sum v_i = 0$. But in that case we can write
$$v = \sum_{i=2}^{2n} v_i(e_i - e_1)$$
and therefore every $v$ orthogonal to $(1,1,\dots, 1)$ belongs to $\mathrm{span}(S)$, as desired.
Note that this doesn't give an explicit set of $n$ vectors, merely shows one exists (although you can probably extract such a set by looking at pairs of vectors of the form $v$ and $v'$ used in this proof).
A: Yes, what you want is possible. Here is a conceptual explanation. Let $S_n$ be the symmetric group acting on vectors by permutations. The matrix $J$ commute with all the matrix of $S_n$. It follows that the eigenspaces of $J$ are invariant subspaces of the $S_n$ action. It is well-known that the action of $S_n$ has just two invariant subspaces, namely the 1-dimensional subspace generated by the vector all whose entries are 1's and its orthogonal complement say $W$. Thus, the action of $S_n$ on $W$ is irreducible hence the $S_n$-orbit of any vector having half of his entries $1$ and the other half $-1$ can not be contained in a subspace of $W$. So these vectors (i.e. the vectors having half its entries as 1 and half as −1)  are a system of generators of $W$. So you can extract from them a basis of the eigenspace relative to $ev = 0$ i.e. a basis of $\ker(J)=W$.
