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Thread: Symmetry of two similar matrices

  1. #1
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    Symmetry of two similar matrices

    Hi!

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

    The rotation matrix:
    $\displaystyle
    P =
    \left[
    \begin{array}{c c c c c c c}
    1 & 0 & \cdots & \cdots & \cdots & 0 & 0 \\
    0 & 0 & \cdots & & & \cdots & \cdots \\
    \cdots & \cdots & cos \theta & \cdots & sin \theta & \cdots & \cdots \\
    \cdots & \cdots & & & & \cdots & \cdots \\
    \cdots & \cdots & -sin \theta & \cdots & cos \theta & \cdots & \cdots \\
    \cdots & \cdots & & & & \cdots & \cdots \\
    0 & 0 & \cdots & \cdots & \cdots & 0 & 1
    \end{array}
    \right]
    $
    where $\displaystyle P_{rr} = P_{ss} = cos \theta, P_{rs} = -P_{sr} = sin \theta, P_{ii} = 1 $ (for $\displaystyle i ~= r, s$) and else $\displaystyle 0$.


    I don't know where to start!

    Help is greatly appreciated!
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  2. #2
    Junior Member
    Joined
    Oct 2009
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    Woho! I produced the
    $\displaystyle B_{rs} = B_{sr} = sin\theta cos\theta (A_{rr} - A{ss}) + (cos^2\theta - sin^2 \theta) A_{rs}$
    simply by multiplying $\displaystyle A$ from both sides with $\displaystyle P^{-1}$ and $\displaystyle P$.

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

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

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

    Hence, B is symmetric when A is symmetric.

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