Today I started studying Rotation matrices and derived a rotation matrix like this:

\begin{equation*} \begin{pmatrix} x & y \end{pmatrix} \begin{pmatrix} \cos(\beta) & \sin(\beta) \\ -\sin(\beta) & \cos(\beta) \end{pmatrix} \end{equation*}

When I was googling for the solution to check if this was ok, I found almost everybody expressed it like this:

\begin{equation*} \begin{pmatrix} \cos(\beta) & -\sin(\beta) \\ \sin(\beta) & \cos(\beta) \end{pmatrix} \begin{pmatrix} x \\ y \end{pmatrix} \end{equation*}

If I'm not mistaken, this leads to the same equations, but is there anything else special about expressing it in these 2 different forms? Has this something to do with Handedness?

  • 2
    $\begingroup$ Your annotations are correct. The handedness can be inferred from the determinant of your rotation matrix (left-hand coordinate system if negative.) As for the different formats, it depends on what type of storage your vectors use; either column or row -wise that you select to pre or post -multiply for your rotations $\endgroup$ – trox Aug 4 '16 at 23:35
  • $\begingroup$ Ok, I see, so as a conclusion: The way I express this matrix operation has nothing to do with handeness (I mean, use row or column matrix). The handeness is given by the determinat. Could you confirm this? $\endgroup$ – Notbad Aug 5 '16 at 11:07
  • $\begingroup$ confirmed as @Ridac asserts $\endgroup$ – trox Aug 5 '16 at 16:09

The difference is that you used a row vector and the other expression uses a column vector. And they are the same if you transpose one of them. \begin{align} \left( \begin{pmatrix} x & y \end{pmatrix} \begin{pmatrix} \cos(\beta) & \sin(\beta) \\ -\sin(\beta) & \cos(\beta) \end{pmatrix} \right)^T &= \begin{pmatrix} \cos(\beta) & \sin(\beta) \\ -\sin(\beta) & \cos(\beta) \end{pmatrix} ^T \begin{pmatrix} x & y \end{pmatrix} ^T \\ &= \begin{pmatrix} \cos(\beta) & -\sin(\beta) \\ \sin(\beta) & \cos(\beta) \end{pmatrix} \begin{pmatrix} x \\ y \end{pmatrix} \end{align}

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