# If $Av = 0$ for a non-zero vector $v$, does the determinant of subrows of A = 0?

Let $A \in \mathbb{Z}^{m \times n}$ be a non-zero matrix where $m \ge n$, and $v \in \mathbb{Z}^{n \times 1}$ be non-zero vector, such that $$Av = \mathbf{0}.$$ Let $\{S_i\}$ be a family of $n$-element sets with elements from $\{1, \dots, m\}.$ Let $A(S_i)$ denote the $n \times n$ matrix formed by the $n$ rows of $A$ having indices from $S_i.$

What do we know about the determinants of the $A(S_i)$ for different $S_i$? Is $\det(A(S_i)) = 0$ for all $i$?

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Hint: Note that $A(S) v = (A v)(S)$.