Prove the solution set to homogeneous equations with this form Let a homogeneous system be given by:
$$\left(\begin{matrix}
0    & \pm1 & \pm1 & \ldots & \pm1 \\
\pm1 & 0    & \pm1 & \ldots & \pm1 \\
\pm1 & \pm1 & 0    & \ldots & \pm1 \\
&&\ldots \\
\pm1 & \pm1 & \pm1    & \ldots & 0 \\
\end{matrix}\right){\bf{x}}={\bf{0}}$$
Where this matrix is $n\times n$, where $n$ is odd, and the sum of each row is zero - that is, there are exactly $(n-1)/2$ ones, $(n-1)/2$ minus ones, and one zero in each row.
I wish to prove that the solution set is the set of vectors $\bf{x}$ with all elements of $\bf{x}$ being equal.
Is there a nice matrix based way of proving this? I think this may be proved using induction, but a non-inductive method would be preferable.
 A: We only have to show that the matrix (let's call it $M$) has rank $n-1$. Then the vectors with identical components are the only solutions, there cannot be any other solutions which are linearly independent of those.
If we can show that one of the diagonal minors is invertible, we are done. We use the fact that the invertibility of $A$ in $\mathbb{Q}^{m\times m}$ follows from the invertibility of the matrix in $\mathbb{F}_2^{m\times m}$. (Assume that $A$ is singular in $\mathbb{Q}^{m\times m}$. Then there must be a vector $v\neq 0$ with integer elements such that $Av=0$. We can assume that there is at least one odd element in this vector, otherwise we could divide all elements by $2$. If we interpret the elements of this vector and the elements of $A$ as elements of $\mathbb{F}_2$, we have obviously found a non-trivial vector for which $Av=0$ in $\mathbb{F}_2^m$. Therefore $A$ is also singular in $\mathbb{F}_2^{m\times m}$) Now we only have to show that the diagonal minors are invertible in $\mathbb{F}_2^{(n-1)\times (n-1)}$ if $n$ is odd. This is easy, because they are all their own matrix inverse.
A: Let me describe a solution which is pretty much equivalent to Reinhard's solution (the main idea is the same) but doesn't involve the determinant.
The question doesn't specify which field we are working for so if we want to try and simplify the problem, we can ask whether the result is true when we work over the simplest field $\mathbb{Z}_2$. Since over $\mathbb{Z}_2$ we have $+1 = -1$, this simplifies the question considerably so let us solve the simple version first.
Call your matrix $A$ and let $B$ be the matrix which has all $1$'s (so we have $A = B - I$). The matrix $B$ is a rank-one matrix so we can analyze it easily. It has an $n-1$ dimensional kernel (consisting of vectors $(x_1,\dots,x_n)^T$ that satisfy $x_1 + \dots + x_n = 0$) and the vector $(1,\dots,1)^T$ is an eigenvector of $B$ which corresponds to the eigenvalue $\lambda  = 1$ (only here we use the fact that $n$ is odd). Hence $\dim \ker(A) = \dim \ker(B - I) = 1$ and we solved the problem for $\mathbb{Z}_2$.
How can we use this to solve the problem over $\mathbb{Q}$? If $n = 1$, the problem is trivial so assume $n > 1$ and that $Ax = 0$ has a solution over $\mathbb{Q}$ with $x_i \neq x_j$. I will show that given such an $x$, we can construct another solution $x' \in \mathbb{Z}^n$ with $Ax' = 0$ such that $x'$ has an even coordinate and an odd coordinate. But then reducing $Ax' = 0$ modulo $2$ we will get a solution over $\mathbb{Z}_2$ which is not $(0,\dots,0)^T$ or $(1,\dots,1)^T$, a contradiction. The reducing modulo $2$ is done by defining $y'_i = x'_i \mod 2$ and $(A')_{ij} = A_{ij} \mod 2$. Since $\operatorname{mod}$ respects addition and multiplication, the equation $Ax' = 0$ over $\mathbb{Z}$ will imply that $A' y' = 0$ over $\mathbb{Z}_2$.
So let $x \in \mathbb{Q}^n$ solve $Ax = 0$ with $x_i \neq x_j$. By multiplying $x$ with an integer, we can assume that $x \in \mathbb{Z}^n$. 
By subtracting $(x_i, \dots, x_i)^T$ from $x$ we can assume that $x_i = 0$. Since $x_i = 0$ it cannot be that all the coordinates in $x$ are odd and since $n > 1$, it cannot be that all the coordinates in $x$ are zero. Consider the following process:


*

*If $x$ has an odd coordinate, we are done.

*If all the coordinates in $x$ are even and some coordinate is non-zero, divide $x$ by $2$ and go to step one.


After finitely many steps, we will get a solution in $\mathbb{Z}^n$ with an even coordinate ($x_i = 0$) and an odd coordinate $\pm 1$, obtaining the required contradiction.

How about other fields? Since our matrix has entries $0,\pm 1$ and the rank of a matrix over a field is the same as the rank of the matrix over the subfield generated by the entries of the matrix, it is enough to investigate the prime fields $\mathbb{Z}_p$ and $\mathbb{Q}$. We have handled $\mathbb{Q}$ and $\mathbb{Z}_2$. For $n = 3$, the matrices are still of rank $n - 1$ over any field. However, for $n = 5$ and $\mathbb{F} = \mathbb{Z}_3$, we can give a counterexample (found using sage):
$$ A = \begin{pmatrix} 0 & 1 & -1 & -1 & 1 \\ 1 & 0 & -1 & 1 & -1 \\ 1 & -1 & 0 & -1 & 1 \\ -1 & 1 & -1 & 0 & 1 \\ 1 & -1 & -1 & 1 & 0 \end{pmatrix}, x = \begin{pmatrix} 1 \\ 0 \\ 2 \\ 1 \\ 0 \end{pmatrix}. $$
Here, $Ax = 0$ while $x \notin \operatorname{span} (1,\dots, 1)^T$.
