0
$\begingroup$

2 x 2 orthogonal matrix $A$ is called proper if detA=1. I know this is a rotation matrix through an angle, and entries of this matrix is composed of $\sin$ and $\cos$.

If you are only given the fact that $2\times 2$ matrix is proper matrix, can I still find out the general form of this matrix? If so, how?

$\endgroup$
  • $\begingroup$ What do you mean by a proper matrix? $\endgroup$ – Pedro Tamaroff Nov 7 '14 at 1:55
  • $\begingroup$ Oops, Real orthogonal matrix is called proper if determiant is 1 . I forgot to mention it $\endgroup$ – Daniel Nov 7 '14 at 1:58
1
$\begingroup$

Suppose that $A = \begin{bmatrix} a & b \\ c & d \end{bmatrix}$ and $A$ is orthogonal.

Then the rows (and columns) are orthogonal and unit length. So $a^2 + b^2 =1$. So that $(a,b)$ lies on the unit circle. Thus $a = \cos(\theta)$ and $b = \sin(\theta)$ for some angle $\theta$. Likewise $a^2+c^2=1$ so $c^2=1-a^2=1-\cos^2(\theta)=\sin^2(\theta)$. Thus $c=\pm \sin(\theta)$. Likewise, $d=\pm \cos(\theta)$.

Suppose that $c = +\sin(\theta)$, then we need $ac+bd=0$ so $\sin(\theta)\cos(\theta)\pm\sin(\theta)\cos(\theta)=0$. Thus $d=-\cos(\theta)$.

Otherwise, $c=-\sin(\theta)$ and we'll need to have $d=\cos(\theta)$.

This leaves us with two options:

$A = \begin{bmatrix} \cos(\theta) & \sin(\theta) \\ \sin(\theta) & -\cos(\theta) \end{bmatrix}$ or $A = \begin{bmatrix} \cos(\theta) & \sin(\theta) \\ -\sin(\theta) & \cos(\theta) \end{bmatrix}$

The first option has $\mathrm{det}(A)=-\cos^2(\theta)-\sin^2(\theta)=-1$ while the second has $\mathrm{det}(A)=\cos^2(\theta)+\sin^2(\theta)=1$.

So every "proper" $2 \times 2$ orthogonal matrix must look like $A = \begin{bmatrix} \cos(\theta) & \sin(\theta) \\ -\sin(\theta) & \cos(\theta) \end{bmatrix}$ for some $\theta$.

$$A = \begin{bmatrix} \cos(\theta) & \sin(\theta) \\ -\sin(\theta) & \cos(\theta) \end{bmatrix} = \begin{bmatrix} \cos(-\theta) & -\sin(-\theta) \\ \sin(-\theta) & \cos(-\theta) \end{bmatrix} = \begin{bmatrix} \sin(\pi/2-\theta) & -\cos(\pi/2-\theta) \\ \cos(\pi/2-\theta) & \sin(\pi/2-\theta) \end{bmatrix} = \mbox{etc.} $$

$\endgroup$
  • $\begingroup$ If you swap cos and sin, does that imply the same result? $\endgroup$ – Daniel Nov 7 '14 at 2:15
  • $\begingroup$ Yes. Keep in mind that $\cos(\theta) = \sin(\pi/2-\theta)$ (and $\sin(\theta) = \cos(\pi/2-\theta)$). Also, since $\sin(-\theta)=-\sin(\theta)$ you can swap out $\theta$ with $-\theta$ and get a matrix with the minus sign in the upper-right hand corner instead of the lower-left hand corner. There are a lot of equivalent ways to write this same matrix. :) $\endgroup$ – Bill Cook Nov 7 '14 at 2:24
  • $\begingroup$ Thank you so much. I really appreciate your sophisticated answer. I $\endgroup$ – Daniel Nov 7 '14 at 2:26
  • $\begingroup$ No problem Daniel. Glad to help! $\endgroup$ – Bill Cook Nov 7 '14 at 2:27

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.