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Math Help - Orthogonal complement

  1. #1
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    Orthogonal complement

    Let V denote the real vector space M_(n,n) (R) of n by n matrices with real entries and let S denote the subspace of all n by n symmetric matrices in V (i.e. such that A=A transpose).

    ---- Determine the orthogonal complement of S in V. (Hint: use a basis of S)

    Thanks..
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  2. #2
    MHF Contributor Drexel28's Avatar
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    Quote Originally Posted by qwerty1234 View Post
    Let V denote the real vector space M_(n,n) (R) of n by n matrices with real entries and let S denote the subspace of all n by n symmetric matrices in V (i.e. such that A=A transpose).

    ---- Determine the orthogonal complement of S in V. (Hint: use a basis of S)

    Thanks..
    The answer is the antisymmetric matrices. Here's a proof: don't peek until you've exhausted all other avenues:
    Spoiler:

    Let \text{Sym}_n\left(\mathbb{R}\right) be the set of all symmetric n\times n matrices and \text{ASym}_n\left(\mathbb{R}\right) the set of all anti-symmetric matrices (that is matrices A with A^\top=-A). I claim that \text{Sym}_n\left(\mathbb{R}\right)^\perp=\text{AS  ym}_n\left(\mathbb{R}\right). Indeed, assuming that you are imposing the inner product on \text{Mat}_n\left(\mathbb{R}\right) by identifying it with \mathbb{R}^{n^2} you can readily prove (if you're really desperate a proof can be gleaned from here) that \left\langle A,B\right\rangle=\text{tr}\left(AB^{\top}\right) from where it's immediate that \text{ASym}_n\left(\mathbb{R}\right)\subseteq\text  {Sym}_n\left(\mathbb{R}\right)^\perp since if
    A\in\text{ASym}_n\left(\mathbb{R}\right),S\in\text  {Sym}_n\left(\mathbb{R}\right) then

    \left\langle A,B\right\rangle=\text{tr}\left(AB^{\top}\right)= \text{tr} \left(-AB\right)=-\text{tr}\left(AB\right)

    but with equal validity

    \left\langle A,B\right\rangle=\left\langle B,A\right\rangle=\text{tr}\left(BA^\top\right)= \text{tr} \left(BA\right)

    and so \left\langle A,B\right\rangle=0. Now, to finish the argument note that every matrix M\in\text{Mat}_n\left(\mathbb{R}\right) may be written as

    \displaystyle M=\underbrace{\frac{M+M^\top}{2}}_{\text{symmetric  }}+\underbrace{\frac{M-M^\top}{2}}_{\text{anti-symmetric}}

    and so \text{Mat}_n\left(\mathbb{R}\right)=\text{Sym}_n \left(\mathbb{R}\right)+\text{ASym}_n\left(\mathbb  {R}\right) that said it's evident that \text{Sym}_n\left(\mathbb{R}\right)\cap\text{ASym}  _n\left(\mathbb{R}\right)=\{\bold{0}\} since if M is in the intersection then M=M^\top=-M. Thus, \text{Mat}_n\left(\mathbb{R}\right)= \text{Sym}_n \left(\mathbb{R}\right)\oplus\text{ASym}_n\left( \mathbb{R} \right) and so

    \text{codim}\left(\text{ASym}_n \mathbb{R}\right)=\dim\text{Sym}_n\left(\mathbb{R}  \right)=\text{codim} \text{Sym}_n\left(\mathbb{R}\right)^\perp

    And so recalling that \text{ASym}_n\left(\mathbb{R}\right)\subseteq\text  {Sym}_n\left(\mathbb{R}\right)^\perp you may conclude with a dimension argument
    Last edited by Drexel28; May 11th 2011 at 08:37 PM.
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  3. #3
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    well a basis for M_n(\mathbb{R}) is \{E_{ij}, 1\leq i,j\leq n\} where each E_{ij} has a 1 in the i,j-th position and 0's elsewhere.

    a basis for S is \{E_{ii}\} \cup \{E_{ij}+E_{ji}, i \neq j\}. considering each of the E_{ij} as standard basis elements in \mathbb{R}^{n^2}, we see any element of S^{\perp} must be orthogonal to each and every one of the basis elements for S.

    in particular, this means that the i,i-th entry of  T \in S^{\perp} must be 0 for all i, and that for i ≠ j, the i,j-th entry of T must be the negative of the j,i-th entry.

    in short, T^{t} = -T, that is, T is anti-symmetric.

    on the other hand, if T is anti-symmetric, it is clear that the entries on the diagonal are 0, so <E_{ii},T> = 0 and the i,j-th entry of T, t_{ij} = -t_{ji}, the j,i-th entry of T, so we have <E_{ij}+E_{ji},T> = 0 therefore, T \in S^{\perp}.
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  4. #4
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    thanks so much, i understand better now!!
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