Results 1 to 5 of 5

Math Help - Linear transformations

  1. #1
    Senior Member
    Joined
    Feb 2008
    Posts
    297

    Linear transformations

    Let T be the rotation in the plane \mathbb{R}^2 through angle \frac{\pi}{3} in the anti-clockwise direction. Find the matrix which represents the linear transformation 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.
    Follow Math Help Forum on Facebook and Google+

  2. #2
    A Plied Mathematician
    Joined
    Jun 2010
    From
    CT, USA
    Posts
    6,318
    Thanks
    4
    Awards
    2
    Look here.
    Follow Math Help Forum on Facebook and Google+

  3. #3
    Senior Member
    Joined
    Feb 2008
    Posts
    297
    Ok thanks, so is that just a general result that should be remembered

    This is what I thought of doing.

    Consider a complex number a+ib

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

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

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

    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}

    =\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; October 7th 2010 at 08:27 PM.
    Follow Math Help Forum on Facebook and Google+

  4. #4
    MHF Contributor

    Joined
    Apr 2005
    Posts
    14,973
    Thanks
    1121
    As to why rotation through angle \theta is given by
    \begin{bmatrix}cos(\theta) & -sin(\theta) \\ sin(\theta) & cos(\theta)\end{bmatrix}

    look at what it does to the "basis vectors"
    \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 (cos(\theta), sin(\theta). The distance from (0, 0) to that point is \sqrt{cos^2(\theta)+ sin^2(\theta)}= 1 and that point lies along a line with slope \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 \theta with the x-axis. That is, it has been rotated through angle \theta.

    \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.
    Follow Math Help Forum on Facebook and Google+

  5. #5
    A Plied Mathematician
    Joined
    Jun 2010
    From
    CT, USA
    Posts
    6,318
    Thanks
    4
    Awards
    2
    acevipa: Yep, it all looks good.
    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. Linear Transformations and the General Linear Group
    Posted in the Advanced Algebra Forum
    Replies: 2
    Last Post: December 26th 2011, 10:50 AM
  2. Linear Map transformations (Linear Algebra)
    Posted in the Algebra Forum
    Replies: 4
    Last Post: October 21st 2011, 09:56 AM
  3. Basic Linear Algebra - Linear Transformations Help
    Posted in the Advanced Algebra Forum
    Replies: 6
    Last Post: December 7th 2010, 03:59 PM
  4. Linear Independence in linear transformations
    Posted in the Advanced Algebra Forum
    Replies: 1
    Last Post: September 17th 2009, 04:22 PM
  5. Replies: 3
    Last Post: June 2nd 2007, 10:08 AM

Search Tags


/mathhelpforum @mathhelpforum