I would like to ask if you may know how to prove this claim or any theorem related:


Given 9 points (x,y,z) lie on unit sphere in 3 dimensional space such that any 4 points are not on the same plane. Let vector a = [1, x, y, z]. Form the 16x9 matrix A = [a1⊗a1, a2⊗a2,.. ,a9⊗a9]. Prove rank of A is 9. (where is the tensor product or the Kronecker product)


I used matlab to confirm this claim, but I wonder if there is any concrete proof for that.
Thank you in advance. 



rank of outer product
I don't have a counterexample, but I do have a proof of something slightly weaker. (This uses the basic argument from [a previous answer of mine](https://mathoverflow.net/questions/136251/on-tensor-products-of-generic-vectors/136256#136256).)


Consider all possible ensembles of $9$ points: $\{(x\_i,y\_i,z\_i)\}\_{i=1}^9\subseteq\mathbb{R}^3$. Then generic ensembles have the property that $V:=\{(x\_i,y\_i,z\_i,x\_iy\_i,y\_iz\_i,z\_ix\_i,x\_i^2,y\_i^2,z\_i^2)\}\_{i=1}^9$ spans $\mathbb{R}^9$. To see this, consider the determinant of the $9\times 9$ matrix whose columns are $V$. This is a polynomial of the coordinates of the $9$ original points. Furthermore, the polynomial is nonzero since there exists an input at which the polynomial is nonzero (based on your MATLAB experiments). As such, the zeros of this polynomial form a set of measure zero, and so almost every input (i.e., ensembles of $9$ points) will yield a nonzero determinant.


