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Rotation matrix between two orthonormal frames

  1. Jul 7, 2015 #1
    I am reading a paper and am stuck on the following snippet.

    Given two orthonormal frames of vectors ##(\bf n1,n2,n3)## and ##(\bf n'1,n'2,n'3)## we can form two matrices ##N= (\bf n1,n2,n3)## and ##N' =(\bf n'1,n'2,n'3)##. In the case of a rigid body, where the two frames are related via rotations and translations only, we can can calculate the rotation matrix between the two frames as:
    ##R = N' N^{T} ##.

    My linear algebra is quite rusty and I am having some trouble understanding why this is true. Thanks for looking.
     
  2. jcsd
  3. Jul 7, 2015 #2

    fzero

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    Let ##\{\mathbf{e}_i\}## be the standard basis ##\mathbf{e}_1 = (1,0,0)^T,\ldots##. Then ##N^T## is the change of basis from ##\{ \mathbf{n}_i \}## to ##\{\mathbf{e}_i\}##, since we can show that ##\mathbf{e}_1 = N^T \mathbf{n}_1##, etc. Similarly, ##N'## is the change of basis from ##\{\mathbf{e}_i\}## to ##\{ \mathbf{n}'_i \}##. Therefore the change of basis from ##\{ \mathbf{n}_i \}## to ##\{ \mathbf{n}'_i \}## is given by the product ##N' N^T##.
     
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