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Math Help - 2-norm is invariant with respect to orthogonal transformations

  1. #1
    Member Mollier's Avatar
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    2-norm is invariant with respect to orthogonal transformations

    Hi,

    my book says that it is easy to show that for all orthogonal Q and Z of appropriate dimensions we have

    ||QAZ||_2 = ||A||_2.

    I'm thinking that Q\in \mathbb{C}^{m\times m}, A\in \mathbb{C}^{m\times n} and Z\in \mathbb{C}^{n\times n}.

    The 2-norm of a matrix is defined as,

    ||A||_2 = \sup_{||x||_2=1}||Ax||_2

    and so

    ||QAZ||_2 = \sup_{||x||_2=1}||(QAZ)x||_2.

    I do not know what to do here.. Any tips are greatly appreciated. Thanks.
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  2. #2
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    Where have you used the fact that Q and Z are orthogonal?
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  3. #3
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    Hint: Orthogonal matrices preserve the (usual) inner product (or equivalently the norm).
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  4. #4
    Member Mollier's Avatar
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    Here's an attempt..

    \begin{aligned}<br />
||QAZ||^2_2 =& \sup_{||x||_2=1}||(QAZ)x||^2_2 \\<br />
                    =& \sup_{||x||_2=1}(QAZx)^*(QAZx) \\<br />
                    =& \sup_{||x||_2=1}(AZx)^*(AZx) \\<br />
                    =& \sup_{||x||_2=1}||(AZx)||^2_2 <br />
\end{aligned}

    Since ||Zx||^2_2=(Zx)^*(Zx)=x^*x=||x||^2_2, we get that

    \sup_{||x||_2=1}||(AZx)||^2_2 = \sup_{||x||_2=1}||(Ax)||^2_2 = ||A||^2_2.

    That last part looks dodgy...
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  5. #5
    MHF Contributor FernandoRevilla's Avatar
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    I don't see clear your third equality. Denote \left\|{\;\;}\right\|_2= \left\|{\;\;}\right\|

    I'd better use:

    \left\|{M}\right\|=\sqrt{\rho(M^*M)}

    Then,

     (QAZ)^*(QAZ)=Z^*A^*Q^*QAZ=Z^*A^*AZ=Z^{-1}(A^*A)Z

    That is,

    (QAZ)^*(QAZ),\;A^*A

    are similar matrices. As a consequence:

    \rho[(QAZ)^*(QAZ)]=\rho(A^*A)

    which implies:

    \left\|{QAZ}\right\|=\left\|{A}\right\|


    Fernando Revilla
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