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I am working on a fun problem. The problem is to solve the following equation: $O_1x_1+O_2x_2=x_1+x_2$, where $x_1, x_2 \in \mathbb{R}^2$ are known and $O_1,O_2 \in \mathbb{O}(2)$ are unknowns. [$\mathbb{O}(2)$ is the group of orthogonal matrices.]

Also, consider the fact that $x_1 \neq \alpha x_2$ where $\alpha \in \mathbb{R}$.

My guess is that the solution is: $$O_1x_1=x_1 \text{ and } O_2x_2=x_2$$ or, $$O_1x_1=x_2 \text{ and } O_2x_2=x_1 \text{ if } \lVert x_1 \rVert = \lVert x_2 \rVert.$$

My attempt is:

$$\lVert O_1x_1+O_2x_2 \rVert^2= \lVert x_1+x_2 \rVert^2 \implies O_2^{\top}O_1x_1=x_1+\alpha x_2^{\perp}.$$ For any vector $d$, the following relationship holds, $$\langle {O_1x_1+O_2x_2 , d} \rangle = \langle {x_1+x_2 , d} \rangle .$$

I could not proceed further after this.

If my guess is correct, then I want an algebraical solution for this. If not, then I need an example to contradict it.

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    $\begingroup$ What is unknown $O_i$ or $x_i$? $\endgroup$
    – V. Moretti
    Jun 1, 2016 at 9:54
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    $\begingroup$ Unless you provide further restrictions, it is easy to find trivial solutions - $O_1=O_2=\mathrm{Id}$, or $x_1=x_2=0$. What sort of statements are you trying to prove about the solution space of the equation? For one, the dichotomy you pose is not true; a counterexample is $$\begin{pmatrix}0&1&0\\1&0&0\\0&0&1\end{pmatrix}\begin{pmatrix}1\\2\\3\end{pmatrix} +\begin{pmatrix}0&1&0\\1&0&0\\0&0&1\end{pmatrix}\begin{pmatrix}2\\1\\4\end{pmatrix} =\begin{pmatrix}1\\2\\3\end{pmatrix} +\begin{pmatrix}2\\1\\4\end{pmatrix}$$ where $O_1x_1$ is neither of $x_1$ or $x_2$. $\endgroup$
    – E.P.
    Jun 1, 2016 at 10:05
  • $\begingroup$ @ValterMoretti thanks. Please check the revised one. $\endgroup$
    – Rajat
    Jun 1, 2016 at 10:11
  • $\begingroup$ @EmilioPisanty thanks. Sorry, I forgot to say that $x_1$ and $x_2$ are linearly independent. I already mentioned that $x_1, x_2 \in \mathbb{R}^2$ $\endgroup$
    – Rajat
    Jun 1, 2016 at 10:13
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    $\begingroup$ This question seems like it would be more at home on Math.SE $\endgroup$ Jun 1, 2016 at 12:09

2 Answers 2

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Here is a counterexample.

$$ \left( \begin{array}{cc} 0 & -1 \\ 1 & 0 \end{array} \right) \left( \begin{array}{c} 1 \\ 0 \end{array} \right) + \left( \begin{array}{cc} 24/25 & 7/25 \\ -7/25 & 24/25 \end{array} \right) \left( \begin{array}{c} 3 \\ 4 \end{array} \right) = \left( \begin{array}{c} 0 \\ 1 \end{array} \right) + \left( \begin{array}{c} 4 \\ 3 \end{array} \right) = \left( \begin{array}{c} 1 \\ 0 \end{array} \right) + \left( \begin{array}{c} 3 \\ 4 \end{array} \right) $$

It clearly satisfies all the conditions. How did I find it? I focused on finding a counterexample with matrices in SO$(2)$, since SO$(2)$ is isomorphic to U$(1)$, that is, the complex numbers of norm $1$. A convenient isomorphism is given by:

$$ \left( \begin{array}{cc} a & b \\ -b & a \end{array} \right) \mapsto a-bi $$

With this isomorphism it is easy to check that

$$ \left( \begin{array}{cc} a & b \\ -b & a \end{array} \right) \left( \begin{array}{c} c \\ d \end{array} \right) = \left( \begin{array}{c} x \\ y \end{array} \right) \Leftrightarrow (a-bi)(c+di)=x+yi $$

And so the question can be rephrased with complex numbers as having $ \alpha x_1 + \beta x_2 = x_1 + x_2$ with $\alpha$ and $\beta$ complex numbers of norm $1$ and $x_1$ and $x_2$ complex numbers which do not lie on a line in the complex plane. I set up this problem with $x_1 = 1$ and $x_2 = 3+4i$ and found by simple manipulation that $\alpha=i$ and $\beta = 24/25 - 7i/25$, which is the solution I gave above.

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Since you are working in $\mathbb{R}^2$, the answer can be seen geometrically: The orthogonal matrices consist of all rotations and all reflections. Furthermore, we know that the dot product (and hence the length of $x+y$) is conserved for all rotations.

First separate angular and length part:

$O_1x_1+O_2x_2=O_1(x_1+O_1^TO_2x_2)$. Since $O_1$ doesn't change the length, you first have to answer for what orthogonal matrices will $x_1+Ox_2$ have the same length as $x_1+x_2$? Well

$$\langle x_1+Ox_2,x_1+Ox_2\rangle=\langle x_1,x_1\rangle + \langle x_2,x_2\rangle+2\langle x_1,Ox_2\rangle$$

But then, $\langle x_1,Ox_2\rangle=\langle x_1,x_2\rangle$, which can only be true for the identity and the reflection along $x_1$, since if $O$ is a rotation, it changes the angle between $x_1$ or $x_2$ and hence the scalar product.

First suppose that $O$ was the identity. Then you only have to solve $\tilde{O}(x_1+x_2)=x_1+x_2$. There is only one rotation which can do that (identity) and one reflection (reflection along $x_1+x_2$).

Now suppose that $O$ was the reflection along $x_1$. Then you have to solve $\tilde{O}(x_1+Ox_2)=x_1+x_2$. Since orthogonal transformations are are rotations and reflections, we have two possibilites:

  • $\tilde{O}$ reflects along an axis in between $x_1+x_2$ and $x_1+Ox_2$. This axis is clearly $x_1$, hence $\tilde{O}(x_1+Ox_2)=x_1+x_2$ and we have the identity.

  • $\tilde{O}$ rotates $x_1+Ox_2$ to $x_1+x_2$. From a picture, you can then see that the result will be a reflection along $x_1+x_2$. If you don't like that, count angles: The angle between $x_1+Ox_2$ and $x_1+x_2$ is twice the angle between $x_1$ and $x_1+x_2$ since we did a reflection along $x_1$. Thus in total, we rotate $x_1$ by twice the angle between $x_1$ and $x_1+x_2$ in one direction (let's assume we rotated clockwise). We first reflect $x_2$ along $x_1$, which is equivalent to rotating it twice the angle between $x_1$ and $x_2$ (this time counterclockwise). Then we rotate it back twice the angle between $x_1$ and $x_1+x_2$ (this time clockwise again), which means that we can just as well rotate $x_2$ by twice the angle between $x_1+x_2$ and $x_2$ (counterclockwise). But this is equivalent to a reflection along $x_1+x_2$!

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  • $\begingroup$ $\langle x_1,Ox_2\rangle=\langle x_1,x_2\rangle$ from this we can only say that: $Ox_2 = x_2+\alpha x_1^{\perp}$ where $\alpha \in \mathbb{R}$ because $O$ can be reflection matrix, and $x_1$ is the corresponding reflection axis. $\endgroup$
    – Rajat
    Jun 1, 2016 at 19:08
  • $\begingroup$ The sign of angle does not matter for the inner product because $\cos \theta = \cos \left( -\theta \right)$. $\endgroup$
    – Rajat
    Jun 1, 2016 at 19:14
  • $\begingroup$ What you say is not quite correct, because $O$ preserves lengths. But you are right that I missed that $O$ could be a reflection along $x_1$. That makes it slightly more complicated. $\endgroup$
    – Martin
    Jun 1, 2016 at 19:22

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