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The statement is as follows: For any ${\bf x}\in R^3$, there exist some ${\bf y}\in R^3$ such that $\begin{bmatrix}x_1 \\ x_2 \\ x_3\end{bmatrix} =\begin{bmatrix}y_1 \\ y_2 \\ y_3\end{bmatrix}\times \begin{bmatrix}y_3 \\ y_1 \\ y_2\end{bmatrix}$ where $\times$ is cross product.

I tried to use a system of 3 variables and 3 equations, where I fix ${\bf x}$, and where $\sin\theta$ depends on the values of ${\bf y}$:

$\begin{bmatrix}x_1 \\ x_2 \\ x_3\end{bmatrix}\cdot \begin{bmatrix}y_1 \\ y_2 \\ y_3\end{bmatrix} =0, \begin{bmatrix}x_1 \\ x_2 \\ x_3\end{bmatrix}\cdot \begin{bmatrix}y_3 \\ y_1 \\ y_2\end{bmatrix}=0$, and finally ${|\bf x}|=|{\bf y}|^2|\sin\theta|$

However, this is not getting me anywhere as there could be some cases where I believe the system is inconsistent. How should I approach this problem?

Extra note: The original problem was to prove this on the neighborhood of ${\bf x} = \begin{bmatrix}0 \\ 0 \\ 1\end{bmatrix}$, but apparently my professor believed that it can be proven for all ${\bf x} \in R^3$

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  • $\begingroup$ What's the $\times$? Cross product? Elementwise product? $\endgroup$ – acarturk Jan 13 at 2:06
  • $\begingroup$ @acarturk Sorry I didn't clarify, it is the cross product $\endgroup$ – LHC2012 Jan 13 at 5:35
  • $\begingroup$ Looks like you’ve already got your answer: the system is inconsistent in some cases, therefore the original statement is false. $\endgroup$ – amd Jan 13 at 22:34
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No, the statement is false.

Counter example: ${\bf x}=<-1,0,0>$. The only solutions for ${\bf y}$ are ${\bf y}=<0,\pm i,0>$, which are not real.

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  • $\begingroup$ How did you find this counterexample? Did you find that the system is inconsistent? $\endgroup$ – LHC2012 Jan 13 at 5:51
  • $\begingroup$ The counter example is still consistent but the solutions are not real. I found it by observing that if $x_1+x_2+x_3<0$, then there are no real solutions for $\bf y$. $\endgroup$ – Pythagoras Jan 13 at 5:57
  • $\begingroup$ How did you come to the conclusion that if $𝑥1+𝑥2+𝑥3<0$, then there is no real solution for ${\bf y}$? I still do not see, why is $y_1^2+y_2^2+y_3^2<y_1\cdot y_2+y_2\cdot y_3+y_3\cdot y_1$ impossible in the real numbers? $\endgroup$ – LHC2012 Jan 13 at 6:06
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    $\begingroup$ $y_1^2+y_2^2+y_3^2-y_1y_2-y_2y_3-y_3y_1=\frac 12[(y_1-y_2)^2+(y_2-y_3)^2+(y_3-y_1)^2]\geq 0$. $\endgroup$ – Pythagoras Jan 13 at 6:11
  • $\begingroup$ Just making sure, the contrapositive is also true right? that is if $x_1^2+x_2^2+x_3^2>=0$, we got a system of $y_1,y_2,y_3$ that is guaranteed to be consistent on $R$ $\endgroup$ – LHC2012 Jan 13 at 6:42
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It appears that you already have the basic answer: your system of equations is inconsistent in some cases, therefore the original statement is false. Pythagoras has given you a counterexample in his answer. Still, it’s interesting to examine this more closely to find the set of vectors for which it is true.

For problems like this one I find it helpful to look at the geometry described by the equations instead of getting bogged down in the algebra. First of all, note that $\mathbf y'=(y_3,y_1,y_2)^T$ can be obtained from $\mathbf y=(y_1,y_2,y_3)^T$ by a rotation $R$ about $\mathbf k=(1,1,1)^T$ through an angle of $2\pi/3$. If $\mathbf y$ doesn’t lie along the rotation axis, then $\mathbf y$, $\mathbf y'$ and $\mathbf k$ form a right-handed system (i.e., $\mathbf y\times\mathbf y'\cdot\mathbf k\gt0$). Since $\mathbf y$, $\mathbf y'$ and $\mathbf x$ are also right-handed, $\mathbf x$ must be on the same side of the plane $x+y+z=0$ as $\mathbf k$, therefore one condition is that $\mathbf x\cdot\mathbf k = x_1+x_2+x_3\gt0$ if $\mathbf x$ is nonzero.

Next, we know that $\mathbf x$ is orthogonal to both $\mathbf y$ and $\mathbf y'$. The second of these conditions can be expressed as $$\mathbf x\cdot\mathbf y' = \mathbf x\cdot R\mathbf y = R^{-1}\mathbf x\cdot\mathbf y = 0.$$ That is, $\mathbf y$ is orthogonal to both $\mathbf x$ and its image under a rotation of $-2\pi/3$ about $\mathbf k$, therefore it’s a multiple of $$\mathbf x\times R^{-1}\mathbf x = (x_1x_2-x_3^2,x_2x_3-x_1^2,x_3x_1-x_2^2)^T.$$ This implies that $$\mathbf y\times\mathbf y' = \lambda^2(x_1+x_2+x_3)(x_1^2+x_2^2+x_3^2-x_1x_2-x_2x_3-x_3x_1)\mathbf x.$$ The quadratic form in the above expression is positive semidefinite, so we once again have the condition $x_1+x_2+x_3\gt0$. We also need $x_1^2+x_2^2+x_3^2-x_1x_2-x_2x_3-x_3x_1\gt0$. This quadratic form vanishes when $x_1=x_2=x_3$, i.e., when $\mathbf x$ is a multiple of $\mathbf k$. Geometrically, this corresponds to the fact that in this case $R^{-1}\mathbf x = \mathbf x$, so there isn’t a unique line defined by their cross product. However, when $\mathbf x$ is a positive multiple of $\mathbf k$, it’s clear that we can take any vector perpendicular to $\mathbf x$ and scale it appropriately to obtain $\mathbf y$.

Finally, then, after simplifying we have $$\mathbf y = \pm(x_1^3+x_2^3+x_3^3-3x_1x_2x_3)^{1/2}(x_1x_2-x_3^2,x_2x_3-x_1^2,x_3x_1-x_2^2)^T$$ when $x_1+x_2+x_3\gt0$ and $\mathbf x\ne(\lambda,\lambda,\lambda)^T$. If, on the other hand, $\mathbf x = (\lambda,\lambda,\lambda)^T$, $\lambda\gt0$, then any vector perpendicular to $\mathbf x$ with length equal to $\left(2\lVert x\rVert/\sqrt3\right)^{1/2}$ will do for $\mathbf y$, and if $\mathbf x=0$, then obviously $\mathbf y=0$ works, although any multiple of $(1,1,1)^T$ also fits the bill. When $x_1+x_2+x_3\le0$ and $\mathbf x\ne0$, there is no solution.

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