TITLE: Largest $n\times m$ matrix with no $n\times (n+1)$ submatrix having $n$-wise independent columns
QUESTION [14 upvotes]: What is the largest $m$ for which an $n \times m$ matrix with the given property that any subset of $n+1$ columns have rank $n$, may contain no subset of $n+1$ columns in which the columns are $n$-wise independent?
In general one can see that the assumptions imply that $m \leq 2n$, but I feel it should be possible to determine $m$ exactly. If $n = 3$ I can verify by hand that the maximum $m$ is $5$. What should it be in general?
Edit:
As some people have pointed to, the question can be generalized to matroids. However in the interest of keeping things concrete, I am willing at this point to accept any answer that exhibits a matrix with entries over any field that improves the general bound $n+2 \leq m \leq 2n$. If the example of Heptagon can be extended to arbitrary $n$, this would for instance show that $2n-1 \leq m \leq 2n$ over the set of realizable matroids.

REPLY [2 votes]: This is just a remark showing that $m=n+3$ is possible. Consider the matrix $$\begin{pmatrix}0&1&1&1&0&0&0\\1&0&1&0&1&0&0\\1&1&0&0&0&1&0\\1&1&1&0&0&0&1\\\end{pmatrix}$$ over the field with two elements $\mathbb{F}_2$.
It has rank four, and it can be checked straightforwardly that any family of five columns contains both a subfamily of four linearly dependent columns and a subfamily of four linearly independent columns. (By the way, this matrix arises as a realization of the dual of the Fano matroid.)

UPD. Several further observations. The equivalent transformations of the rows of matrices do not change the properties mentioned in the question. Therefore, we can assume without loss of generality that, if $A$ is a matrix as in the question, then $A=(I|A')$ with $I$ the identity $n\times n$ matrix. Note that every column of $A'$ must contain a zero since otherwise the matrix formed out of this column and the columns of $I$ has all its $n\times n$ submatrices non-singular.
If it happens that $A'$ contains a $k\times(k+1)$ submatrix (with row indices in the set $R$ and column indices in the set $C$, say) of rank less than $k$, then the matrix formed by the $(n+1)$ columns of $A$ with indexes in $\{1,\ldots,n\}\cup C\setminus R$ has rank less than $n$, a contradiction. Therefore, every $k\times(k+1)$ submatrix of $A'$ has rank $k$.
Over the field $\mathbb{F}_2$, the latter condition (applied with $k=1,2$ only) says that, if $m\geqslant n+3$, then every column of $A'$ contains at most one zero, and actually it contains exactly one zero if we use the result of the former paragraph. This means that $A$ must follow the pattern as in the initial answer above, and we see that $m\leqslant n+2$ unless $n=4$.
Finally, we get a complete answer for this problem over $\mathbb{F}_2$. (Here, I denote by $m_i$ the maximal possible value of $m$ corresponding to $n=i$.) Analyzing the small values of $n$ separately and taking into account what is said above, we get $m_1=2$, $m_2=3$, $m_3=5$, $m_4=7$, and $m_i=i+2$ for $i>4$.
My conjecture is that the answer is the same over all fields except that $m_4=6$ in characteristic different from two. (In other words, the example with $m=n+3$ is possible only for $n=4$ and in characteristic two.) But a proof of this conjecture seems to require a much more trickier combinatorial analysis if it is plausible at all.