# Simple vector statement, how to prove?

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 $${\bf x} \in R^3$$

• What's the $\times$? Cross product? Elementwise product? – acarturk Jan 13 at 2:06
• @acarturk Sorry I didn't clarify, it is the cross product – LHC2012 Jan 13 at 5:35
• Looks like you’ve already got your answer: the system is inconsistent in some cases, therefore the original statement is false. – amd Jan 13 at 22:34

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.

• How did you find this counterexample? Did you find that the system is inconsistent? – LHC2012 Jan 13 at 5:51
• 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 $\bf y$. – Pythagoras Jan 13 at 5:57
• 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? – LHC2012 Jan 13 at 6:06
• $y_1^2+y_2^2+y_3^2-y_1y_2-y_2y_3-y_3y_1=\frac 12[(y_1-y_2)^2+(y_2-y_3)^2+(y_3-y_1)^2]\geq 0$. – Pythagoras Jan 13 at 6:11
• 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 $R$ – LHC2012 Jan 13 at 6:42

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.