Suppose that $x\_1,\ldots,x\_n$ are $n$ vectors in $\mathbb{R}^m$ (where $m<n\leq m^2$) such that any subset of $m$ of them are linearly independent (i.e., they are "generic"). Now, form the $m^2\times n$ matrix 
$A = [x\_1\otimes x\_1, x\_2\otimes x\_2,\cdots,x\_n\otimes x\_n]$ (where $\otimes$ is the tensor product or the [Kronecker product](http://en.wikipedia.org/wiki/Kronecker_product))


Is there a nice way of showing that this matrix is full rank (i.e., rank$(A)=n$)?




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Edit: (after the great answers below)


* The statement, as I asked, was shown to be false by Clement de Seguins Pazzi with a very nice argument.
* The statement that I had *hoped* to ask is true and the details can be found in Dustin Mixon's answers



On tensor products of "generic" vectors
The result is not true and the best upper-bound on $n$ to make it work happens to 
be $2m-1$. 


Assume that $n<2m$. Let $(t\_1,\dots,t\_n)$ be a family of real numbers
such that $\sum\_{k=1}^n t\_k \,x\_k \otimes x\_k=0$. Assume that some $t\_k$ is non-zero. 
Denote by N the number of indices $k$ such that $t\_k \neq 0$. 
Obviously $N>m$. Without loss of generality, we may assume that $t\_1,\dots,t\_m$ are all non-zero, and we rewrite the above equality as 
$\sum\_{k=1}^m t\_k x\_k \otimes x\_k=-\sum\_{k=m+1}^n t\_k x\_k \otimes x\_k$. 
As $(x\_1,\dots,x\_m)$ and $(x\_{m+1},\dots,x\_n)$ are both linearly independent, 
one sees that the rank of $\sum\_{k=1}^m t\_k x\_k \otimes x\_k$ is $m$, whereas the one of
$\sum\_{k=m+1}^n t\_k\,x\_k \otimes x\_k$ is at most $n-m<m$. This is a contradiction. 


To see that $2m-1$ is an optimal upper-bound, note first that
the problem may be entirely restated in terms of families of symmetric rank $1$ matrices. One takes a family $(X\_i)\_{1 \leq i \leq n}$ of vectors of $\mathbb{R}^m$
in which every subfamily with $m$ vectors is a basis of $\mathbb{R}^m$, 
and one tries to find the rank of the family $(X\_i X\_i^T)\_{1 \leq i \leq n}$ of
rank $1$ symmetric matrices.
Now, for $a \in \mathbb{R}$, consider the vector $X(a)=(a^k)\_{0 \leq k \leq n-1}$. 
Then, for every $n$-tuple $(a\_1,\dots,a\_n)$ of real numbers, the vectors 
$X(a\_1),\dots,X(a\_n)$ are linearly independent; however, the matrices $X(a)X(a)^T$
all belong to the space of real $m \times m$ Hankel matrices, which has dimension $2m-1$, 
and hence a family consisting of such matrices has rank at most $2m-1$. 


