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Thread: Symmetric matrix

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

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

    If not, when is it true?

    (I'm trying to show that $\displaystyle <u, v>=u^t M v$ forms an inner product, but as far as I can see $\displaystyle <u, v> \neq <v, u>$...)
    Last edited by Swlabr; Jan 28th 2010 at 01:04 PM.
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  2. #2
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    It is $\displaystyle u^t \times M\times v,$ isn't it? This is a scalar (a 1-1 matrix).

    Therefore it is equal to its transpose, and $\displaystyle u^tMv=(u^tMv)^t=v^tM^tu=v^tMu$ whenever $\displaystyle M$ 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 $\displaystyle u^t \times M\times v,$ isn't it? This is a scalar (a 1-1 matrix).

    Therefore it is equal to its transpose, and $\displaystyle u^tMv=(u^tMv)^t=v^tM^tu=v^tMu$ whenever $\displaystyle M$ 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 $\displaystyle u=v$?
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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 $\displaystyle u=v$?
    no. it depends on $\displaystyle u$ and $\displaystyle M.$ for example if $\displaystyle u=e_1 = [1 \ 0 \ 0 \cdots \ 0]^T$ and $\displaystyle g_{11} < 0,$ then $\displaystyle u^TMu=g_{11} < 0.$
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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 $\displaystyle u$ and $\displaystyle M.$ for example if $\displaystyle u=e_1 = [1 \ 0 \ 0 \cdots \ 0]^T$ and $\displaystyle g_{11} < 0,$ then $\displaystyle u^TMu=g_{11} < 0.$
    Hmm. Well, I've been asked to prove that the assignment,

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

    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; Jan 29th 2010 at 12:05 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,

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

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