TITLE: A tricky problem about matrices with no $\{-1,0,1\}$ vector in their kernel
QUESTION [9 upvotes]: A Hankel matrix is a square matrix in which each ascending skew-diagonal from left to right is constant. Let us call a matrix partial Hankel if it is the first $m<n$ rows of some $n$ by $n$ Hankel matrix. I was posed this question by a mathematician a few years ago and I have been completely stuck on it ever since.   Let us assume all our partial Hankel matrices have elements which are chosen from $\{-1,1\}$.

Does there exist an $m$ by $n$ partial Hankel matrix $M$ such that it has $m < n/2$ rows and there is no non-zero vector $v\in\{-1,0,1\}^n$ in its kernel?

Such matrices certainly do exist for non-Hankel $m$ by $n$ $\{-1,1\}$-matrices.
Other that writing computer code to exhaustively explore small matrices I am not sure how to approach this question. 
I would be very grateful for any ideas at all.

REPLY [2 votes]: This is a comment on Dierk Bormann's answer with roughly the same estimate.
The case $m< c\frac{n}{\log n}$ can be done using the pigeonhole principle.
For every vector $\mathbf{x}\in\{0,1\}^n$, take the $m$-vector $M\mathbf{x}$.
Every coordinate of $M\mathbf{x}$ is some integer in the interval $[-n,n]$. So there are $2^n$ choices for the vector $\mathbf{x}$, and at most $(2n+1)^m$ possible results for $M\mathbf{x}$. If $2^n>(2n+1)^m$ then there are some vectors $\mathbf{x},\mathbf{y}\in \{0,1\}^n$ such that 
$M\mathbf{x}=M\mathbf{y}$. Then $v=\mathbf{x}-\mathbf{y}$ satisfies the requred properties.
So a suitable vector $v$ exists if $2^n>(2n+1)^m$, or equivalently $m<\frac{n}{\log_2(2n+1)}$.
This works for all $m\times n$ matrices with entries from a fixed set of integers, so it does not use the special form of $M$.

Another approach is applying Minkowski's theorem. I did not have time to work it out for this case, it requires some unpleasant estimates, but it is clear to me that the same $m<c\frac{n}{\log n}$ can be achieved, and perhaps it is possible improve the estimate using the stucture of $M$.
The idea is as follows.
We need a vector $v$ from the kernel of $M$, such that every $v_i$ is an integer, lying in the interval $(-2,2)$.
The kernel $\mathrm{ker}\,M = \{x\in\mathbb{R}^n: Mx=0\}$ is a subspace of $\mathbb{R}^n$ with dimension $d$; obviosly $d\ge n-m$.
The integer vectors in this subspace form a lattice. We need an upper bound on the $d$-volume of the lattice parallelepiped.
We need a suitable lower bound on the $d$-volume of the set 
$K = (-2,2)^n \cap \mathrm{ker}\,M$ which is convex and symmetric wrt. the origin.
If the ratio of the two volumes is high enoough, we can apply Minkowski's theorem.