Describing Rotations Geometrically

Question

Could someone give an idea on how to complete this? Suppose I have a 2x2 matrix, where all entries but the bottom left are sqrt(2)/2. The bottom left however, is -sqrt(2)/2. With that being said, how I started it was by plugging in the values; for instance, the top left entry of the matrix is cos(sqrt(2)/2), bottom right is the same thing, etc. But I have a feeling this is wrong; any help here?

Answer 1

A rotation matrix in $\mathbb{R}^2$ through an angle $\theta$ is given by

$R_\theta=\left (\begin{matrix} \cos{\theta} & -\sin{\theta} \\ \sin{\theta} & cos{\theta} \end{matrix} \right )$

For your example, it seems that you want to find $\theta$ such that

$\left (\begin{matrix} \cos{\theta} & -\sin{\theta} \\ \sin{\theta} & \cos{\theta} \end{matrix} \right ) = \left (\begin{matrix} \sqrt{2}\over 2 & \sqrt{2}\over 2 \\ -\sqrt{2}\over 2 & \sqrt{2}\over 2 \end{matrix} \right )$

which is definitely easy to solve by looking at a unit circle.

To "draw" it, you need to multiply some points in $\mathbb{R}^2$ by the rotation matrix. For example, if you wanted to rotate $(x, y) = (0,1)$ through your given matrix, it would look like this:

$\left (\begin{matrix} \sqrt{2}\over 2 & \sqrt{2}\over 2 \\ -\sqrt{2}\over 2 & \sqrt{2}\over 2 \end{matrix} \right ) \left ( \begin{matrix} 0 \\ 1 \end{matrix} \right ) = \left ( \begin{matrix} \sqrt{2} \over 2 \\ \sqrt{2} \over 2 \end{matrix} \right ) $

So your new coordinates would be $(\frac{\sqrt{2}}{2},\frac{\sqrt{2}}{2})$.

Does that help?