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Math Help - Show that A is an orthogonal matrix

  1. #1
    MHF Contributor alexmahone's Avatar
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    Show that A is an orthogonal matrix

    A=(I-B)^{-1}(I+B)

    Show that A is always an orthogonal matrix (ie A^{T}A=I) if B^{T}=-B.

    My working:

    A^{T}=(I+B)^{T}[(I-B)^{-1}]^{T}

    =(I+B)^{T}[(I-B)^{T}]^{-1}

    =(I^T+B^T)(I^T-B^T)^{-1}

    =(I-B)(I+B)^{-1}

    I don't seem to be getting anywhere. Any suggestions?
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  2. #2
    A Plied Mathematician
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    You've correctly computed A^{T}. Now form both products AA^{T} and A^{T}A. For the AA^{T} product, note that (I-B) and (I+B) commute (just multiply it out both ways, and you'll see why). That will get you the result you want for that product. As for the A^{T}A version, which you also should prove, that takes a bit more doing, because you really want elements with the inverses to commute also. Thinking...
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  3. #3
    MHF Contributor alexmahone's Avatar
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    Quote Originally Posted by Ackbeet View Post
    You've correctly computed A^{T}. Now form both products AA^{T} and A^{T}A. For the AA^{T} product, note that (I-B) and (I+B) commute (just multiply it out both ways, and you'll see why). That will get you the result you want for that product. As for the A^{T}A version, which you also should prove, that takes a bit more doing, because you really want elements with the inverses to commute also. Thinking...
    (I+B)(I-B)=(I-B)(I+B)

    Taking inverses of both sides,

    (I-B)^{-1}(I+B)^{-1}=(I+B)^{-1}(I-B)^{-1} ---------- (1)

    Thus, A^{T}A=(I-B)(I+B)^{-1}(I-B)^{-1}(I+B)

    =(I-B)(I-B)^{-1}(I+B)^{-1}(I+B) [Using (1)]

    =I
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  4. #4
    A Plied Mathematician
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    Very nice! I think you're done.
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  5. #5
    MHF Contributor FernandoRevilla's Avatar
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    To complete the proof, it is necessary to say that (I-B)^{-1} always exists if B^t=-B . Equivalently, \det (I-B)\neq 0 or \lambda=1 is not an eigenvalue of B.

    In fact, the eigenvalues of a real skew symmetric are of the form \alpha i\;(\alpha\in\mathbb{R}) .
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  6. #6
    A Plied Mathematician
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    Quote Originally Posted by FernandoRevilla View Post
    To complete the proof, it is necessary to say that (I-B)^{-1} always exists if B^t=-B . Equivalently, \det (I-B)\neq 0 or \lambda=1 is not an eigenvalue of B.

    In fact, the eigenvalues of a real skew symmetric are of the form \alpha i\;(\alpha\in\mathbb{R}) .
    Very nice, but isn't the invertibility of (I-B) assumed by the definition of A? Maybe you need to prove that (I+B) is invertible. Certainly your proof would be sufficient to show that as well, I think.
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  7. #7
    MHF Contributor FernandoRevilla's Avatar
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    Quote Originally Posted by Ackbeet View Post
    Very nice, but isn't the invertibility of (I-B) assumed by the definition of A?

    Well, perhaps, but I don't see it very clear.


    Maybe you need to prove that (I+B) is invertible. Certainly your proof would be sufficient to show that as well, I think.

    Right, in the same way: -1\neq \alpha i for all \alpha \in \mathbb{R} .
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