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Thread: Orthogonal problem

  1. #1
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    Orthogonal

    If A and B are both orthogonal nxn matrices.

    Can anyone show That AB is orthogonal...

    Thanks
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  2. #2
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    Orthogonal problem

    If A and B are both orthogonal nxn matrices.

    Can anyone show That AB is orthogonal...iv been stuck on this question for a while now

    Thanks
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  3. #3
    Grand Panjandrum
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    Quote Originally Posted by dopi
    If A and B are both orthogonal nxn matrices.

    Can anyone show That AB is orthogonal...iv been stuck on this question for a while now

    Thanks
    An $\displaystyle n\times n$ $\displaystyle A$ is orthogonal iff

    $\displaystyle AA^T=I$

    Now suppose $\displaystyle A$ and $\displaystyle B$ be orthogonal, and let
    $\displaystyle C=AB$. Now $\displaystyle C^T=B^TA^T$, so:

    $\displaystyle
    CC^T=ABB^TA^T=AIA^T=AA^T=I
    $

    Hence $\displaystyle C$ is orthogonal.

    RonL
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  4. #4
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    Quote Originally Posted by CaptainBlack
    An $\displaystyle n\times n$ $\displaystyle A$ is orthogonal iff

    $\displaystyle AA^T=I$

    Now suppose $\displaystyle A$ and $\displaystyle B$ be orthogonal, and let
    $\displaystyle C=AB$. Now $\displaystyle C^T=B^TA^T$, so:

    $\displaystyle
    CC^T=ABB^TA^T=AIA^T=AA^T=I
    $

    Hence $\displaystyle C$ is orthogonal.

    RonL

    Say if you add A and B (A+B)...would that be Orthognal?
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  5. #5
    Grand Panjandrum
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    Quote Originally Posted by dopi
    Say if you add A and B (A+B)...would that be Orthognal?
    Not generally

    RonL
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  6. #6
    TD!
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    From the definition, we have for A and B:

    $\displaystyle
    \begin{array}{l}
    AA^{ - 1} = A^{ - 1} A = I \\
    BB^{ - 1} = B^{ - 1} B = I \\
    \end{array}
    $

    $\displaystyle
    \left( {AB} \right)\left( {AB} \right)^{ - 1} = ABB^{ - 1} A^{ - 1} = AIA^{ - 1} = AA^{ - 1} = I
    $
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  7. #7
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    Don't make duplicate posts

    Don't post the same question in two different fora.

    I have merged these two because they both have responses.

    RonL
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  8. #8
    Grand Panjandrum
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    Quote Originally Posted by TD!
    From the definition, we have for A and B:

    $\displaystyle
    \begin{array}{l}
    AA^{ - 1} = A^{ - 1} A = I \\
    BB^{ - 1} = B^{ - 1} B = I \\
    \end{array}
    $

    $\displaystyle
    \left( {AB} \right)\left( {AB} \right)^{ - 1} = ABB^{ - 1} A^{ - 1} = AIA^{ - 1} = AA^{ - 1} = I
    $
    This is true of any non-singlar square matrices $\displaystyle A$ and $\displaystyle B$!

    RonL
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