Why $x\in\ker A$ implies $x_i-x_j=\lambda \det A_{ij}$? Suppose that $A$ is a real matrix with $n-2$ linearly independent rows and $n$ columns adding up to $0$. I can show that for any $x=(x_1,\dotsc,x_n)\in\ker A$ (that is, any $x\in\mathbb R^n$ satisfying $Ax=0$), there exists $\lambda\in\mathbb R$ such that
  $$ x_i-x_j=\lambda\det A_{ij},\quad 1\le i<j\le n, $$
where $A_{ij}$ is the matrix obtained from $A$ by removing the $i$th and the $j$th columns. This seems a nice fact to me, but my proof is rather ugly. Is there a simple, nice, "symmetric" (non-discriminating) proof?
As a side remark, the full-rank assumption is essential: consider, for instance, the situation where $n=3$, $A=(0\ 0\ 0)$, and $x=(1,2,3)^t\in\ker A$.
 A: The statement in question is wrong. The correct one should be
$$
x_i-x_j=(-1)^{i-j}\lambda\,\det A_{ij},\quad 1\le i<j\le n.\tag{1}
$$
We may assume that $\mathbf x\in\ker A$ is not a scalar multiple of $\mathbf 1$ (the all-one vector), otherwise the statement is trivial. We will see that the corrected statement is just a consequence of Cramer's rule.
Let $\mathbf a_1,\ldots,\mathbf a_n$ be the columns of $A$. Since $\mathbf1\in\ker A$, we get $\mathbf a_j=-\sum_{i\ne j}\mathbf a_i$. Therefore, if $\mathbf x\in\ker A$ and $y_i=x_i-x_j$, then $0=\sum_ix_i\mathbf a_i=\sum_{i\ne j}y_i\mathbf a_i$. So, if we let $\mathbf z^T=(y_1,\ldots,y_{j-1},y_{j+1},\ldots,y_n)$ and let $B$ be the matrix obtained by inserting $\mathbf z^T$ above the $j$-th row of $\begin{bmatrix}\mathbf a_1&\cdots&\mathbf a_{j-1}&\mathbf a_{j+1}&\cdots&\mathbf a_n\end{bmatrix}$ (so that $\mathbf z^T$ becomes the $j$-th row of the new matrix), then
$$
B\mathbf z=\|\mathbf z\|^2\mathbf e_j,\tag{2}
$$
where $\mathbf e_j$ is the $j$-th vector in the standard basis of $\mathbf R^{n-1}$.
Note that $\mathbf z\ne0$ because $\mathbf x$ does not lie in the span of $\mathbf 1$. Also, as the sum of all columns of $A$ is zero, if we remove a single column from $A$, the column rank and in turn the row rank are preserved. So, $BB^T$ and in turn $B$ are invertible. 
Apply Cramer's rule on $(2)$, we get
$$
z_i=\frac{(-1)^{i+j}\|\mathbf y\|^2 A_{ij}}{\det B}.\tag{3}
$$
In other words, when $i<j$,
$$
x_i-x_j=y_i=z_i=\frac{(-1)^{i+j}\|\mathbf y\|^2 A_{ij}}{\det B}.\tag{4}
$$
And when $i>j$,
$$
x_j-x_i=-y_i=-z_{i-1}=\frac{(-1)^{i+j}\|\mathbf y\|^2 A_{ij}}{\det B}.\tag{5}
$$
Hence the conclusion follows.
A: I totally ran into a wall down below, but I'm sure something like this can be used ... I'll leave it here as is for a while to contemplate.
The desired condition is linear, i.e., it certainly holds for a subspace:
If $\lambda$ works for $x$ and $\lambda'$ works for $x'$ then $c\lambda+c'\lambda'$ works for $cx+c'x'$.
As the columns add up to $0$, the all-ones vector $e=(1,\ldots,1)^T$ is in the kernel, and for this the desired condition holds with $\lambda=0$.
As the kernel is two-dimensional, there is another vector $z=(z_1,\ldots,z_n)^T$ in the kernel that is linearly independent from the all-ones vector. To compete the proof it suffices to show that the desird property holds for $z$ with a suitable $\lambda$.
Let $B$ be the matrix $A$ with two additional rows added on top, formed by $z^T$ and $e^T$.
Then $B$ has full rank, hence non-zero determinant.
We can compute $\det B$ by expanding the first row (i.e., $z^T$):
$$ \det B= \sum_{i=1}^n(-1)^{i-1}z_i\det B_i$$
where $B_i$ is the matrix obtained from removing row $1$ and column $i$.
We can expand $\det B_i$ again at its first row (i.e., $e^T$):
$$ \det B_i = \sum_{j=1}^{n-1}(-1)^{j-1}\det B_{i,j}$$
where $B_{i,j}$ is the matrix obtained from $B_i$ by removing its first row and its $j$th column.
In other words, $$B_{i,j}=\begin{cases}A_{j,i}&j<i\\A_{i,j+1}&j\ge i\end{cases}$$
That makes
$$\begin{align}\det &B=  \sum_{i=1}^n(-1)^{i-1}z_i\sum_{j=1}^{n-1}(-1)^{j-1}\det B_{i,j}\\
&=\sum_{i=1}^n(-1)^{i-1}z_i\left(\sum_{j=1}^{i-1}(-1)^{j-1}\det A_{j,i}+\sum_{j=i}^{n-1}(-1)^{j-1}\det A_{i,j+1}\right)\\
&=\sum_{1\le i<j\le n}\det A_{i,j}\cdot\left((-1)^{i+j-2}z_j+(-1)^{i+j-3}z_i\right)\\
&=\sum_{1\le i<j\le n}(-1)^{i+1}(z_j-z_i)\det A_{i,j}\\
\end{align}$$
Oops ...
