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Math Help - Verifying that a matrix is a rotation matrix

  1. #1
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    Verifying that a matrix is a rotation matrix

    Hi!

    How do you, properly, verify that a given matrix is a rotation matrix?

    Consider the following matrix P:
    <br />
P =<br />
 \left[<br />
\begin{array}{c c c c c c c}<br />
1   & 0   & \cdots  & \cdots  & \cdots & 0 & 0 \\<br />
0   & 0   & \cdots &  &  & \cdots & \cdots \\<br />
\cdots & \cdots  & cos \theta & \cdots & sin \theta & \cdots & \cdots \\<br />
\cdots & \cdots  &  &  &   & \cdots & \cdots \\<br />
\cdots & \cdots  & -sin \theta & \cdots & cos \theta & \cdots & \cdots \\<br />
\cdots & \cdots &  &  &  & \cdots & \cdots \\<br />
0 & 0 & \cdots  & \cdots & \cdots & 0 & 1<br />
\end{array}<br />
\right]<br />
    where P_{rr} = P_{ss} = cos \theta, P_{rs} = -P_{sr} = sin \theta, P_{ii} = 1 (for i ~= r, s) and else 0.

    It's essentially an identity matrix with that familiar 2x2 rotation matrix setup sort of in the middle there. But mathematically, what conditions must I show are true to establish that this is a rotation matrix? I would think orthogonality is one of them?

    Thank you in advance!

    (sorry for the missing \cdots, I exceeded the limit of latex code)
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  2. #2
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    Just check it is orthogonal and that each row has its norm equal to 1. That won't be very troublesome
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  3. #3
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    Thank you!

    Regarding checking that it is orthogonal, the best way is to show that PP^{T}=P^{T}P=I?
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  4. #4
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    Yes in that case it is easy to verify that PP^T=I.
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  5. #5
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    I'm sorry to bother you with more of these simple questions, but when verifying PP^T=I, is it ok notation-wise to omit the larger part of P and only show calculations on a smaller part containing the cos and sin elements? Then show that they either go to 1 or 0 and thus PP^T=I.
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  6. #6
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    Well that depends on the corrector... So it's hard to answer you!

    But even the more detailled version of the answer is not that bad: let's say PP^T=(a_{ij})_{0<i,j<n+1}

    If i or j\neq r,s then for any 1\leq k\leq n,\ P_{ik}P_{jk}=1\Leftrightarrow i=k=j, otherwise it is zero, hence a_{ij}=\sum_{k=1}^nP_{ik}P_{jk}=\delta_{ij}

    Now i,j\in\{r,s\}.\ a_{rr}=a_{ss}=\cos^2\theta +\sin^2\theta=1 and a_{rs}=a_{sr}=-\sin\theta\cos\theta + \cos\theta\sin\theta=0 ... I guess it is evident with the form of theese lines/rows.
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