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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.

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  • $\begingroup$ This might help: mathoverflow.net/questions/136251/… $\endgroup$
    – Dustin G. Mixon
    Commented Jul 14, 2013 at 9:28
  • $\begingroup$ Thank everybody for help. The original problem of this question is: Given any 9 points in R3 that lie on a unit sphere such that any 4 of them are not on the same plane. prove the linear combination of 9 points (x,y,z,xy,yz,zx, x^2,y^2,z^2) is full rank. Based on the counter example, the condition that these 9 points lie on a sphere is necessary. I could only prove this question in R2 when all the points lie on unit circle using trigonometry. I guess this problem can be generalized for higher dimensional space $\endgroup$
    – thomeou
    Commented Jul 15, 2013 at 1:20
  • $\begingroup$ Is this homework? Or for a class? $\endgroup$
    – Dustin G. Mixon
    Commented Jul 15, 2013 at 8:31
  • $\begingroup$ I am doing a small research. Related to your previous post. I could rephrase my question in this form: vector a = [ 1, x, y, z]' in R4 such that x^2 + y^2 + z^2 = 1; Given a_1, a_2,.., a_n vector (9<n<=16) such that any subset of 4 of them are linearly independent and not stay on the same hyperplane. Then matrix A = [a_1\otimes a_1, a_2\otimes a_2,\cdots,a_n\otimes a_n] has rank 9. I observe that if x^2 + y^2 + z^2 != 1 then maximum rank of A could be 10 which is the upper bound $\endgroup$
    – thomeou
    Commented Jul 15, 2013 at 9:16

3 Answers 3

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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.)

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.

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  • $\begingroup$ thank you Dustin, I don't really get your answer yet. I will look though the paper you recommended $\endgroup$
    – thomeou
    Commented Jul 15, 2013 at 1:22
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Suppose you have found $9$ points in $\mathbb{R}^3$ such that no $4$ points lie in the same plane, as required. Apply an isometry such that one point coincides with the origin $(0,0,0)$. Then the determinant of the corresponding $9\times 9$ matrix is zero, since there is a zero column. This should be a counterexample. Generically however the determinant will be nonzero.

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  • $\begingroup$ Generically, in what sense? $\endgroup$
    – Felix Goldberg
    Commented Jul 16, 2013 at 11:30
  • $\begingroup$ If I added the condition that all the point must be on the sphere and none 4 of them are on the same plane. is there anyway to prove the determinant of matrix A=[a1 .... a9] is not zero,where a_i = [x, y, z, xy, yz, zx,x^2, y^2, z^2] $\endgroup$
    – thomeou
    Commented Jul 16, 2013 at 11:31
  • $\begingroup$ @Felix: "Generically" was just refering to Dustin's answer, which was the first one (for the "older" question). $\endgroup$
    – Dietrich Burde
    Commented Jul 16, 2013 at 15:13
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Here is an easy counterexample: take $9$ general points in the cone $xy=z^2$. Then your points are contained in a vector space of codimension $1$.

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