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I asked this question here a fortnight ago and answered it myself after pondering on it. Answers are still very much welcome and appreciated there, my answer is just the best I could come up with. I took the vector $\mathbf{X}=(x_1,x_2,x_3)$, and the related $\mathbf{X^2}=(x_1^2,x_2^2,x_3^2)$, where the $x_t\in\Bbb R/\{0\}$ are not equal.

I could then obtain $\mathbf{V}=\mathbf{X^2}\times\mathbf{X}$ for the simple solution that $c$ solves $$\mathbf{V}\cdot(\mathbf{1}c-\mathbf{Y})=0$$

For clarity $\mathbf{1}$, $c$ and $\mathbf{Y}$ are defined in the question linked at the top of this question - that's the context of this related question

I think that $\mathbf{V}$ is injective, that is, no two $\mathbf{X}$ map to the same $\mathbf{V}$, but I have no idea how to prove this, the algebra seems to go round in circles.

Help would be appreciated - thanks in advance.

Rhys Hughes
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    Seems to me that $(1, 1, x)$ all map to the zero vector. – RghtHndSd Jan 24 '20 at 15:43
  • Apologies. Forgot to mention that the $x_t$ can't be equal as $\mathbf{X}$ is using three different $x$-coordinates. Also I typo'd writing out $\mathbf{X^2}$. Whoops. – Rhys Hughes Jan 24 '20 at 15:50
  • If you have a formula for $V$ in terms of $X$, how could it be non-unique? What is the relevance of $Y$ and $c$? They just seem to appear from nowhere. – timur Mar 25 '20 at 03:33
  • @timur seems I fail at proofreading. I edited my question for clarity. Sorry – Rhys Hughes Mar 26 '20 at 01:34
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    Can you please flesh out exactly what you need here since your original question contain many details that are not relevant here. How does $Y$ depend on $X$? – timur Mar 26 '20 at 03:57
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    $(x_1,y_1), (x_2,y_2), (x_3,y_3)$ are points on a quadratic $Q$ with $Q(0)=c$, $$\mathbf{X}=(x_1,x_2,x_3)$$ $$\mathbf{Y}=(y_1,y_2,y_3)$$ $$\mathbf{1}=(1,1,1)$$ The linked question establishes a method of determining $Q$ from three points, first using the formula $$(\mathbf{X^2}\times \mathbf{X})\cdot(\mathbf{1}c-\mathbf{Y})=0$$ What this question simply asks is if the vector $\mathbf{V}=\mathbf{X^2}\times \mathbf{X}$ can ever be represented by $\mathbf{V}=\mathbf{(X')^2}\times \mathbf{X'}, \mathbf{X'}\neq\mathbf{X}$ (the next step for me is being able to find $\mathbf{X}$ from $\mathbf{V}$) – Rhys Hughes Mar 26 '20 at 13:29

1 Answers1

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Let ${\bf i, j, k}$ be the standard unit vectors of the space $\Bbb R^3$. Namely, ${\bf i}=(1,0,0)$, ${\bf j}=(0,1,0)$, and ${\bf k}=(0,0,1)$. Then $\bf{X^2}\times\bf{X}$ formally equals to a determinant $\left|\begin{matrix} \bf{i} & \bf{j} & \bf{k}\\ x_1 & x_2 & x_3\\ x_1^2 & x_2^2 & x_3^2 \end{matrix}\right|$. In particular, if $x_3=1$ then ${\bf V}=(x_2-x_2^2, x_1^2-x_1, x_1x_2(x_2-x_1))$. The first two coordinates of $\bf{V}$ are preserved by substitutions $x_1\to 1-x_1$ and $x_2\to 1-x_2$. So to show non-injectivity of $\bf{V}$ it suffices to find $x_1\ne x_2$ distinct also from $1$ and $1/2$ such that $x_1x_2(x_2-x_1)=(1-x_1)(1-x_2)(1-x_2-(1-x_1))$ that is $-x_1x_2=(1-x_1)(1-x_2)$. It suffices to have $x_2=\tfrac{1-x_1}{1-2x_1}$. For instance, we can pick $x_1=\tfrac 32$ and $x_2=\tfrac 14$. Then we have that ${\bf V}\left(\tfrac 32, \tfrac 14,1\right)= {\bf V}\left(-\tfrac 12, \tfrac 34,1\right)$.

PS. Even in $\Bbb R^2$ a quadratic is determined by five its points $(x_i, y_i)$, provided no four of them are collinear (see, for instance, [KK, 2.4.11]). In this case the equation of the quadratic is $$\left|\begin{matrix} x^2 & xy & y^2 & x & y & 1\\ x_1^2 & x_1y_1 & y_1^2 & x_1 & y_1 & 1\\ x_2^2 & x_2y_2 & y_2^2 & x_2 & y_2 & 1\\ x_3^2 & x_3y_3 & y_3^2 & x_3 & y_3 & 1\\ x_4^2 & x_4y_4 & y_4^2 & x_4 & y_4 & 1\\ x_5^2 & x_5y_5 & y_5^2 & x_5 & y_5 & 1 \end{matrix}\right|=0.$$

The quadratic is degenerated iff three of the given points are collinear.

References

[KK] Granino Korn, Theresa Korn Mathematical Handbook for scientists and engineers, 2nd edition, McGraw Hill, 1968 (Russian translation, Moskow, “Nauka”, 1973).

Alex Ravsky
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