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Math Help - Symmetric matrix

  1. #1
    MHF Contributor Swlabr's Avatar
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    Symmetric matrix

    Let M=(g_{i, j}) by a symmetric matrix. Then is it true in general that u^t M v = v^t M u for u and v column vectors?

    If not, when is it true?

    (I'm trying to show that <u, v>=u^t M v forms an inner product, but as far as I can see <u, v> \neq <v, u>...)
    Last edited by Swlabr; January 28th 2010 at 02:04 PM.
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  2. #2
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    It is u^t \times M\times v, isn't it? This is a scalar (a 1-1 matrix).

    Therefore it is equal to its transpose, and u^tMv=(u^tMv)^t=v^tM^tu=v^tMu whenever M is symmetric.
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  3. #3
    MHF Contributor Swlabr's Avatar
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    Quote Originally Posted by clic-clac View Post
    It is u^t \times M\times v, isn't it? This is a scalar (a 1-1 matrix).

    Therefore it is equal to its transpose, and u^tMv=(u^tMv)^t=v^tM^tu=v^tMu whenever M is symmetric.
    Thanks - I was thinking it gave a column vector for some reason.

    I do have one more question though: does this always give us a positive value if u=v?
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  4. #4
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    Quote Originally Posted by Swlabr View Post

    does this always give us a positive value if u=v?
    no. it depends on u and M. for example if u=e_1 = [1 \ 0 \ 0 \cdots \ 0]^T and g_{11} < 0, then u^TMu=g_{11} < 0.
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  5. #5
    MHF Contributor Swlabr's Avatar
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    Quote Originally Posted by NonCommAlg View Post
    no. it depends on u and M. for example if u=e_1 = [1 \ 0 \ 0 \cdots \ 0]^T and g_{11} < 0, then u^TMu=g_{11} < 0.
    Hmm. Well, I've been asked to prove that the assignment,

    <\cdot, \cdot>: T_pM \times T_pM \rightarrow \mathbb{R}, (u, v) \mapsto u^t (g_{i, j}) v^t is an inner product, where g_{i, j} = e_i . e_j and the e_i are the basis vectors for T_pM, the tangent space of a manifold at a point p.

    So I need to prove that it is positive definite, but I can't see why this holds. I suppose the point to use would be that elements on the diagonal are all (strictly) positive, but I don't see how to use this.
    Last edited by Swlabr; January 29th 2010 at 01:05 AM.
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  6. #6
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    Quote Originally Posted by Swlabr View Post
    I've been asked to prove that the assignment,

    <\cdot, \cdot>: T_pM \times T_pM \rightarrow \mathbb{R}, (u, v) \mapsto u^t (g_{i, j}) v^t is an inner product, where g_{i, j} = e_i . e_j and the e_i are the basis vectors for T_pM, the tangent space of a manifold at a point p.

    So I need to prove that it is positive definite
    If the matrix M has entries e_i.e_j, and the vector u has components u_1,\ldots,u_n, then u^{\textsc t}Mu = \textstyle\sum_i\sum_ju_i(e_i.e_j)u_j = \Bigl(\sum_iu_ie_i\Bigr).\Bigl(\sum_ju_je_j\Bigr) \geqslant0. So M is positive semi-definite.
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