Construction of a matrix over $ \{-1,0,1\} $ Let  $ Z=(z_{ij})  $ be a  $ (n,n)$-matrix, for which:


*

*$ z_{ij} \in \mathbb{R}; $

*$  z_{ij}= -z_{ji} $ for $ i,j=1, \dots , n; $

*$ \sum_{j=1}^n z_{ij} = 0 $ and $ \sum_{j=1}^n |z_{ij} |>0 $ for $ i=1, \dots , n. $


Please help me to prove that
There exist a  $ (n,n)$-matrix  $ H=(h_{ij}),   $ for which: 


*

*$ h_{ij} \in \{-1,0, 1\}; $

*$ h_{ij}= -h_{ji} $ for $ i,j=1, \dots , n; $

*$ \sum_{j=1}^n h_{ij} = 0 $ and $ \sum_{j=1}^n |h_{ij} |= 2 $ for $ i=1, \dots , n; $

*$ h_{ij}=1 \Rightarrow z_{ij} >0. $

 A: We can build these matrices recursively.
For $n=2$ we define:
$ M_2 := 
\begin{bmatrix}
   -1  &  1 \\
    1  &  -1 \\
\end{bmatrix}
$
Suppose we have defined $M_{n-1}$. We claim that $z_{n-1 n-1}$  element (botom-right element of this matrice) is nonzero. We replace this element by $0$. Hence $z_{n-1 n-1}$ was $\neq 0$, now sum in the last ($n-1$)-th column is $-1$ or $1$.
case A) If it is $-1$, we add the last ($n$-th) row : $\begin{bmatrix} 0 & \ldots & 0 & 1 -1   \end{bmatrix}$ and the last ($n$-th) column:
$\begin{bmatrix} 0 \\ \ldots \\ 0 \\ 1 \\-1   \end{bmatrix}$ .
case B) If it is $1$, we add the last ($n$-th) row : $\begin{bmatrix} 0 & \ldots & 0 & -1 & 1   \end{bmatrix}$ and the last ($n$-th) column:
$\begin{bmatrix} 0 \\ \ldots \\ 0 \\ -1 \\1   \end{bmatrix}$ .
And we can see that $z_{n n}$ element is always $\neq 0$ and the summation rules are preserved. Hence this way we define $M_n$ matrix.
For given matrix $Z$ whe have always the same numer of $-1$-s and $1$ in each column (3rd condition).
And also in matrix $M_n$. Number of $0$-s in $M_n$ column  is always greather or equal than numer of $-1$, $1$-s in $Z$.
Hence we can permute each column of $M_n$ in such way we have always $z_{i j} = 1 $ when $h_{i j} > 0 $.
