Each column has at most $s$ zeros, the matrix has rank $\leq s+1$?

Question

Each column has at most $s$ zeros, the $n\times n$ matrix has rank $\leq s+1$?

Is this true? Can we prove this by induction on $n$ or $r$?

Answer 1

This answers the revised post. Let $I$ be the $n\times n$ identity matrix. To make the matrix $A$, add $2$ to every entry of $I$. The matrix $A$ has full rank. For its determinant is congruent to $1$ modulo $2$.

The matrix $A$ has at most $0$ zeros in each column. Thus $A$ has rank $\gt s+1$ for all $n\ge 2$.