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?

  • $\begingroup$ 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

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 $\theta$.

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

  • $\begingroup$ 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

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