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Math Help - Symmetry of two similar matrices

  1. #1
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    Symmetry of two similar matrices

    Hi!

    Assume B = P^TAP, where P is the rotation matrix (below) with angle \theta and position r, s. Show that B is symmetric when A is, and that we then also have: B_{rs} = B_{sr} = sin\theta cos\theta (A_{rr} - A{ss}) + (cos^2\theta - sin^2 \theta) A_{rs}.

    The rotation matrix:
    <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.


    I don't know where to start!

    Help is greatly appreciated!
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  2. #2
    Junior Member
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    Woho! I produced the
    B_{rs} = B_{sr} = sin\theta cos\theta (A_{rr} - A{ss}) + (cos^2\theta - sin^2 \theta) A_{rs}
    simply by multiplying A from both sides with P^{-1} and P.

    So what's left is to show that B is symmetric when A is... Suggestions on that?
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  3. #3
    Junior Member
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    I think I got it!

    If A is symmetric, then A = A^T. If B is to be symmetric, then B = B^T must also be true. Hence I need to show that B^T = B when A is assumed symmetric.

    Assume A is symmetric.
    We have: B = P^TAP. We then get:

    B^T = (P^TAP)^T = P^TA^TP^{TT} = P^TAP = B

    Hence, B is symmetric when A is symmetric.

    Did I do it correctly?
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  4. #4
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    Yep, that seems perfect
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