Let $M$ be a $m\times n$ matrix with entries in a field $F$ and $R_{1}, R_{2}, \ldots, R_{m}$ be its rows. We also assume that null space of $M$ contains a non-zero element.
Let $v = \begin{bmatrix} v_{1} & v_{2} & \ldots & v_{n} \end{bmatrix}$ be a non-zero element of null space of the matrix $M$. If the row space of $M$ contains $\begin{bmatrix} 1 & 1 & \ldots & 1 \end{bmatrix}$, i.e., $\sum_{i=1}^{k} \mu_{i}R_{i} = \begin{bmatrix} 1 & 1 & \ldots & 1 \end{bmatrix} $ then it easy to show that
$$ \sum_{j=1}^{n} v_{j} = \begin{bmatrix} 1 & 1 & \ldots & 1 \end{bmatrix} \times v = \sum_{i=1}^{k} \mu_{i} R_{i} \times F = \sum_{i=1}^{k} \mu_{i} \big( R_{i} \times F \big) = 0.$$
I want to show that the converse: if the row space of the matrix $M$ does not have $\begin{bmatrix} 1 & 1 & \ldots & 1 \end{bmatrix}$, then there exists a non-zero element $v = \begin{bmatrix} v_{1} & v_{2} & \ldots & v_{n} \end{bmatrix}$ in the null space of $M$ such that $\sum_{j=1}^{n} v_{j} \neq 0$. Since the null space of a matrix is the intersection (or linear combination) of null spaces of its rows $R_{i}$s, I thought this would not be hard to prove. However, my attempts have proved futile so far.
