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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?

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Yes, that is exactly what I'm using in the book! I'm just confused as to how to draw it; I asked in another question about projection, so the problem with me is that I don't quite understand the actual drawing itself...if that makes any sense. Perhaps you could get me started; the theta respectively for the top left entry would be pi/4, the bottom right would be pi/4 too, the top right entry would be pi/4, and the bottom left entry would be 5pi/4. Thanks! –  Linear Algebra Failure Mar 20 '12 at 6:17
    
Your $\theta$ cannot change between the entries (Some can't be $\pi \over 4$ with others being $5\pi \over 4$). $\theta = \frac{\pi}{4}$ means you are rotating by 45 degrees counterclockwise. –  Tyler Mar 20 '12 at 6:20
    
Oh, 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! –  Linear Algebra Failure Mar 20 '12 at 6:27

1 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?

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Ohh, I think so. In my case, I used (x,y) = (1,0) for instance. So my starting line would be at the coordinate (1,0), and would rotate to coordinates (sqrt(2)/2, (-sqrt(2)/2)), which would be a line in the 4th quadrant, exactly 45 degrees, and going counter clockwise, correct? Does that mean I essentially "pick" where my starting coordinates are? And thank you so much for your help. If you have time, perhaps you could help me with projection; 0 idea on that too. –  Linear Algebra Failure Mar 20 '12 at 6:43
    
@LinearAlgebraFailure: Well, I don't know what exactly your book is having your draw, but that is the basic idea. Given any point, and an angle, you can find the rotation matrix, multiply the point (as a vector) by the rotation matrix to find the new, rotated point. It's just convention that we choose counterclockwise rotations. As for your projection question, it seems you've gotten a few answers (probably better than I could do! :P) –  Tyler Mar 20 '12 at 6:57
    
@LinearAlgebraFailure: There are no "starting coordinates" built into the rotation matrix. The rotation matrix has to work for all possible "starting coordinates". –  Ted Mar 20 '12 at 7:32

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