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I found this comic:

XKCD comic showing a 90°-rotation matrix applied to a column vector equals that column vector drawn sideways

But I can't understand the humor because I can't understand how trig functions affect matrix multiplication. Can someone please explain?

4 Answers 4

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The matrix $\left[\begin{align} \cos(\theta) & \sin(\theta) \\ -\sin(\theta) & \cos(\theta) \end{align} \right]$ when it acts on a vector it rotates the vector by $\theta$ in clockwise direction. Hence when $\theta = 90^{\circ}$, it rotates the vector $\left[\begin{align} a_1 \\ a_2 \end{align} \right]$ from vertical to horizontal clockwise.

  • 0
    Oops, missed that *clockwise*.2011-06-03
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Well, the rotation matrix they write there is

$ \left( \begin{array}{cc} 0 & 1 \\ -1 & 0 \\ \end{array} \right) $

If you multiply a vector $\left( \begin{array}{c} a_{1} \\ a_{2} \\ \end{array} \right)$ by this matrix then you end up with $\left( \begin{array}{c} a_{2} \\ -a_{1} \\ \end{array} \right)$ If you draw a picture in the $xy$ plane connecting each of $(a_{1},a_{2})$ and $(a_{2},-a_{1})$ with the origin, it will be clear that the latter is a 90 degree rotation of the former.

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That's a $-90$-degree rotation matrix.

  • 1
    Wow! A highly improbable case of two events occurring simultaneously! Your answer and mine show the same "answered 2 secs ago"!2011-06-03
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Because...

enter image description here

Sorry couldn't resist when it was bumped.