I'm struggling with a problem for quite some time and unfortunately I haven't found the solution yet.. I would appreciate any insights..
Consider the finite set of all vectors of dimension $n$ with exactly $k$ positive ones and $n-k$ zeroes which I denote as $S_{n,k}$. The cardinality of this set is thus, $|S_{n,k}|=\binom{n}{k}$.
Remark: $k<\frac{n}{2}$
Given $n-1$ linearly independent vectors $u_1,...,u_{n-1}$ from $S_{n,k}$, which form the matrix $A$ (i.e. the rows of $A$ are $u_1,...,u_{n-1}$)
What is the minimal number of independent vectors remaining in $S_{n,k}$ for any such $A$?
For example, in case $A$ has a column of zeroes (say $i$), then there are exactly $\binom{n-1}{k-1}$ independent vectors: Those which has $1$ in the i'th entry.
I argue that for any $A$ containing $n-1$ independent vectors from $S_{n,k}$ there are at least $\binom{n-1}{k-1}$ independent vectors to $A$ remaining in $S_{n,k}$. That is, any one of these vectors will increase the rank of $A$ to $n$.
Thanks
It is not true. Consider the case $n=4,k=3$. There are $4$ vectors in $S_{4,3}$. Any set of three of them is independent and can form $A$. You would claim that there are ${3 \choose 2}=3$ vectors left in $S_{4,3}$ that we could add to complete a basis, but there is only $1$.