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?
Describing Rotations Geometrically
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0Oh, you're right. Oops, sorry about that. So this leads me to my next question, how would you draw this geometrically? Would I draw a line to let's say, pi/4, then rotate it somewhere? To me, the main question is, where is my starting line? Thanks! – 2012-03-20
1 Answers
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?
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0@LinearAlgebraFailure: There are no "starting coordinates" built into the rotation matrix. The rotation matrix has to work for all possible "starting coordinates". – 2012-03-20