Largest $n\times m$ matrix with no $n\times (n+1)$ submatrix having $n$-wise independent columns 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.
 A: Here is a partial answer that ends with a nice characterization of 
admissible families in cardinality $n+2$. My current guess is that $m$
is not always equal to $n+2$, and that the results below will be helpful
in constructing a counterexample.
Notation In the sequel, an admissible family of vectors is one that satisfies
your conditions and ${\cal B}=\lbrace e_1,e_2,\ldots,e_n \rbrace $ denotes a basis of the vector space.
Also $x,y,z$ are arbitrary vectors, with coordinates $(x_k),(y_k),(z_k)$ in $\cal B$ (for example
$x=\sum_{i=k}^n x_k e_k$, etc). The supports $X,Y,Z$ for vectors $x,y,z$ are defined by
$X=\lbrace k | x_k\neq 0 \rbrace$, etc.  Note that for a finite family of vectors is admissible
iff every $(n+1)$-subset has both an independent $n$-subset (which we call an
$n$-base)  and a dependent $n$-subset (which we call a $n$-nonbase).
So a family is admissible iff it has enough $n$-bases and enough $n$-nonbases.
For $t\in \lbrace 1,2,\ldots, n\rbrace$, we have
$$
{\sf span}(({\cal B}\setminus \lbrace e_t \rbrace)\cup \lbrace x \rbrace)=
\left\lbrace\begin{array}{lcll}
{\sf span}(({\cal B}\setminus \lbrace e_t \rbrace) &\text{if} \ t\not\in X& \text{so it's a nonbase in this case} \\
{\mathbb K}^n  &\text{if} \ t\in X& \text{so it's a base in this case} \\
\end{array}\right.\tag{1}
$$
where the second line follows from the fact that $e_t$ is a linear combination
of $x$ and the $e_k$ for $k\neq t$ (in fact $e_t=\frac{x-\sum_{k\in X,k\neq t}x_ke_k}{x_t}$).
to have an $n$-nonbase in $\cal B\cup\lbrace x \rbrace$ we need at least one $t\not\in X$.
It follows immediately that
Lemma 1 : Admissibility criterion for $n+1$-families The family $F_{n+1}=\cal B\cup\lbrace x \rbrace $
is admissible iff $X\neq \lbrace 1,2,\ldots, n\rbrace$.
Next, we have :
Theorem : Admissibility criterion for $n+2$-families The 
family $F_{n+2}=\lbrace e_1,e_2,\ldots,e_n,x,y \rbrace$ is admissible iff 
$X\neq \lbrace 1,2,\ldots, n\rbrace$,$Y\neq \lbrace 1,2,\ldots, n\rbrace$,
$X\cup Y=\lbrace 1,2,\ldots, n\rbrace$, and $X\cap Y$ is either empty or can be partitioned into
parts of size at least $2$ on which the ratio $\frac{y_k}{x_k}$ is constant.
Proof of theorem. Let us show the "only if" direction : suppose that $F_{n+2}$ is admissible. Arguing as 
in lemma 1, to have an $n$-nonbase in $\cal B\cup\lbrace x \rbrace$ we need 
$X'=\lbrace 1,2,\ldots, n\rbrace \setminus X$ to be nonempty. Similarly
$Y'=\lbrace 1,2,\ldots, n\rbrace \setminus Y$ is nonempty. Now if $X'\cap Y'$ were nonempty, say $i\in X'\cap Y'$,
then the $n+1$-family $({\cal B})\setminus \lbrace e_i \rbrace)\cup \lbrace x,y \rbrace$ would not span $e_i$, 
and so could not contain any base, contradicting admissibility. So $X'\cap Y'=\emptyset$, in other 
words $X\cup Y=\lbrace 1,2,\ldots, n\rbrace$.  If $X\cap Y=\emptyset$, we are done. So suppose
$X\cap Y\neq \emptyset$ and let $i\in X\cap Y$. Let $G=F_{n+2}\setminus \lbrace e_i \rbrace$. By the admissibility
hypothesis, we have a $g\in G$ such that $H=G\setminus \lbrace g \rbrace$ is an $n$-nonbase. Because of the second
line in (1), we have $g\not\in\lbrace x,y\rbrace$. So $g=e_j$ for some $j\in\lbrace 1,2,\ldots, n\rbrace, j \neq i$.
If $j\not\in Y$, then $e_i=\frac{y-\sum{k\in Y,k\neq i}y_ke_k}{y_i}\in {\sf span}(H)$ and
and hence $e_j=\frac{x-\sum{k\in X,k\neq j}x_ke_k}{x_j}\in {\sf span}(H)$ contradicting 
the fact that $H$ is a nonbase. So we must have $j\in Y$. Interchanging the roles of $X$ and $Y$,
we have $j\in X$ also. Then ${\sf span}(H)$ contains all the $e_k$ for $k\not\in\lbrace i,j\rbrace$, but
also $x^{\sim}=x_ie_i+x_je_j=x-\sum_{k\in X,k\neq i,k\neq j}x_ke_k$ and 
$y^{\sim}=y_ie_i+y_je_j=y-\sum_{k\in Y,k\neq i,k\neq j}y_ke_k$. As $H$ is a nonbase, $x^{\sim}$ and
$y^{\sim}$ are proportional vectors, so $\frac{y_j}{x_j}=\frac{y_i}{x_i}$. Since for any $i$ we can find
a suitable $j$, this is clearly the partition condition formulated above. 
The "if" direction is only a verification "in reverse" of the work we have just done on the
"only if" direction. 
A: 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.
