Let $A$ be an $n \times m$ matrix with each entry equal to $\pm1,0$ such that every column has exactly one $+1$ and exactly one $-1$. Which one of the following is true?
Show that 1. is the correct alternative.
On
Hint: Let $A= (a_{ij}) \in Mat_{n\times m} (\mathbb{Z})$. Note that $$rank\begin{bmatrix} a_{11}&a_{12}&\ldots&a_{1m}\\ a_{21}&a_{22}&\ldots&a_{2m}\\ *&*&*&*\\ a_{n1}&a_{n2}&\ldots&a_{nm}\\ \end{bmatrix} = rank \begin{bmatrix} \sum_{k=1}^n a_{k1}& \sum_{k=1}^n a_{k2}&\ldots&\sum_{k=1}^n a_{km}\\ a_{21}&a_{22}&\ldots&a_{2m}\\ *&*&*&*\\ a_{n1}&a_{n2}&\ldots&a_{nm}\\ \end{bmatrix}$$ Or, $$rank\begin{bmatrix} a_{11}&a_{12}&\ldots&a_{1m}\\ a_{21}&a_{22}&\ldots&a_{2m}\\ *&*&*&*\\ a_{n1}&a_{n2}&\ldots&a_{nm}\\ \end{bmatrix} = rank \begin{bmatrix} 0& 0&\ldots&0\\ a_{21}&a_{22}&\ldots&a_{2m}\\ *&*&*&*\\ a_{n1}&a_{n2}&\ldots&a_{nm}\\ \end{bmatrix}$$ Thus $rank (A) \leq n-1$.
The rank is the dimension of the image of $A$. You can verify, that the image lies in the orthogonal complement of $\mathrm{span}(\begin{pmatrix} 1 \\ \vdots \\ 1 \end{pmatrix})$, which has dimension $n-1$.