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

  • $\begingroup$ 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

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.

| cite | improve this answer | |
  • $\begingroup$ 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
  • 1
    $\begingroup$ $y_1^2+y_2^2+y_3^2-y_1y_2-y_2y_3-y_3y_1=\frac 12[(y_1-y_2)^2+(y_2-y_3)^2+(y_3-y_1)^2]\geq 0$. $\endgroup$ – 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

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.

| cite | improve this answer | |

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.