Unique linear combination in matrix with skew-symmetric condition Let $A$ be an $n\times n$ matrix with real entries such that the numbers in each column sum to $0$, and $a_{ij}\in\{0,1\}$ for all $i\neq j$, and $a_{ij}=0\leftrightarrow a_{ji}=1$ for all $i\neq j$. 
So, $A$ is non-invertible, and some linear combination of columns is equal to the $\textbf{0}$ vector. Is the linear combination necessarily unique (up to constant factor)?
 A: This answer is affirmative.
You are essentially asking if $0$ is a simple eigenvalue of $A$. Let $B=\frac1nA+I$. Then $B$ is entrywise nonnegative and its column sums are all equal to $1$. In other words, $B$ is a row-stochastic matrix. Furthermore, by the given conditions on $A$, we have

($\ast$): $(b_{ij},b_{ji})=(\frac1n,0)$ or $(0,\frac1n)$ for every off-diagonal entry $b_{ij}$.

Now the problem boils down to asking whether $1$ is a simple eigenvalue of $B$. 
Suppose $B$ is reducible. By permuting the rows and columns of $B$, we may assume that $B$ is block upper triangular and its first diagonal block is irreducible.
Note that condition ($\ast$) still holds under permutation-similarity. Yet, this means every column sum of all but the first diagonal blocks is strictly less than $1$. In other words, the induced $1$-norms of all but the first diagonal blocks are strictly less than $1$. Since the spectral radius of a matrix is bounded above by any matrix norm, it follows that $1$ is not an eigenvalue of every diagonal block other than the first.
The first diagonal block, however, is an irreducible stochastic matrix. Therefore, by Perron-Frobenius theorem for nonnegative irreducible matrices, $1$ is a simple eigenvalue of it, and hence $1$ is a simple eigenvalue of $B$.
