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Thread: Linear transformations

  1. #1
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    Linear transformations

    Let $\displaystyle T$ be the rotation in the plane $\displaystyle \mathbb{R}^2$ through angle $\displaystyle \frac{\pi}{3}$ in the anti-clockwise direction. Find the matrix which represents the linear transformation $\displaystyle T$.

    Not too sure how to start this question off. The only thing I can think of doing for this is to use complex numbers.

    If someone could give me a clue on how to start off this question and I'll try and do it myself.
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  2. #2
    A Plied Mathematician
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    Look here.
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  3. #3
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    Ok thanks, so is that just a general result that should be remembered

    This is what I thought of doing.

    Consider a complex number $\displaystyle a+ib$

    Multiply it by $\displaystyle \cos{\frac{\pi}{3}}+i\sin{\frac{\pi}{3}}$

    $\displaystyle \Rightarrow (a+ib)(\frac{1}{2}+\frac{\sqrt{3}}{{2}}i)$

    $\displaystyle =\frac{1}{2}a-\frac{\sqrt{3}}{2}b+i(\frac{\sqrt{3}}{2}a+\frac{1} {2}b)$

    $\displaystyle T\begin{pmatrix}a\\ b\end{pmatrix}=\begin{pmatrix}\frac{1}{2}a-\frac{\sqrt{3}}{2}b\\ \frac{\sqrt{3}}{2}a+\frac{1}{2}b\end{pmatrix}$

    $\displaystyle =\begin{pmatrix}\frac{1}{2}-\frac{\sqrt{3}}{2}\\ \frac{\sqrt{3}}{2}+\frac{1}{2}\end{pmatrix}\begin{ pmatrix}a\\b\end{pmatrix}$

    Is this correct?
    Last edited by acevipa; Oct 7th 2010 at 08:27 PM.
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  4. #4
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    As to why rotation through angle $\displaystyle \theta$ is given by
    $\displaystyle \begin{bmatrix}cos(\theta) & -sin(\theta) \\ sin(\theta) & cos(\theta)\end{bmatrix}$

    look at what it does to the "basis vectors"
    $\displaystyle \begin{bmatrix}cos(\theta) & -sin(\theta) \\ sin(\theta) & cos(\theta)\end{bmatrix}\begin{bmatrix}1 \\ 0\end{bmatrix}= \begin{bmatrix}cos(\theta) & sin(\theta)\end{bmatrix}$
    That is, (1, 0) is mapped to the point $\displaystyle (cos(\theta), sin(\theta)$. The distance from (0, 0) to that point is $\displaystyle \sqrt{cos^2(\theta)+ sin^2(\theta)}= 1$ and that point lies along a line with slope $\displaystyle \frac{sin(\theta)}{cos(\theta)}= tan(\theta)$. Since the slope of a line is the tangent of the angle the line makes with the x-axis, that says that (1, 0) has been mapped to a point on the line that makes angle $\displaystyle \theta$ with the x-axis. That is, it has been rotated through angle $\displaystyle \theta$.

    $\displaystyle \begin{bmatrix}cos(\theta) & -sin(\theta) \\ sin(\theta) & cos(\theta)\end{bmatrix}\begin{bmatrix}0 \\ 1\end{bmatrix}= \begin{bmatrix}-sin(\theta) & cos(\theta)\end{bmatrix}$ tells us the same thing.
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  5. #5
    A Plied Mathematician
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    acevipa: Yep, it all looks good.
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