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Math Help - (n - 1) dimensional submanifold of the manifold R^n

  1. #1
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    (n - 1) dimensional submanifold of the manifold R^n

    Let A be a symmetric n \times n matrix over \mathbb{R}.
    Let 0 \neq b \in \mathbb{R}.


    Show that the surface M = \{x\in \mathbb{R}^n \mid x^T A x = b\} is an (n - 1) dimensional submanifold of the manifold \mathbb{R}^n.


    Please, help!
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  2. #2
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    Re: (n - 1) dimensional submanifold of the manifold R^n

    The differential of the smooth function f: R^n \rightarrow R, f: x \mapsto x^TAx-b is
    f_*: R^n \rightarrow R, f_*: v \mapsto 2x^TAv
    Since b \ne 0, x^tA never vanishes on the locus f=0 so f_* has full rank.
    That is, all points of M are regular points. Hence M is a n-1 dimensional sub-manifold.
    Last edited by xxp9; December 19th 2012 at 07:22 PM.
    Thanks from vercammen
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  3. #3
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    Re: (n - 1) dimensional submanifold of the manifold R^n

    Thank you!!!
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