# Maximum number of $0$-$1$ vectors with a given rank

Let $$k\ge2$$. The maximum number of $$0$$-$$1$$ (column) vectors of length $$2k-1$$ which make a rank $$k$$ matrix with no zero row nor two identical rows is $$2^{k-1}+1$$. (The rank is over the rationals.)

I have a computer-based proof for this statement, but I'm looking for a computer-free argument.

A construction for the lower bound: take the matrix of rank $$k-1$$ with $$2^{k-1}$$ columns consisting of all $$\{0,1\}$$-combinations of the vectors $$e_1+e_2, e_3+e_4 \dots, e_{2k-3}+e_{2k-2}$$. Then add the column $$e_1+e_3+e_5+\dots+e_{2k-1}$$, which will increase the rank by $$1$$ and make all the rows distinct.
Edit: If the rank was computed over $$\mathbb{Z}_2$$, the matrix with $$2^k$$ columns consisting of all $$\{0,1\}$$-combinations of the $$k$$ vectors above also has rank $$k$$ and no zero row or two identical rows:
$$\begin{pmatrix} 0&1&1&0\\ 0&1&0&1\\ 0&0&1&1 \end{pmatrix}$$