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Thread: Dual Space

  1. #1
    Member kezman's Avatar
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    Dual Space

    Let $\displaystyle V = M_n(k)$ Vector space of n X n Matrices and $\displaystyle S \subset V$ Subspace of Simetric Matrices. Calculate a base for $\displaystyle S^o$.
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  2. #2
    Super Member Rebesques's Avatar
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    If $\displaystyle (E_{ij})$ are the basis matrices for $\displaystyle M_n$, then you know a basis for the dual $\displaystyle M_n^*$; It consists of the $\displaystyle n^2$ maps

    $\displaystyle \phi_{ij}:M_n\rightarrow \mathbb{R}$ defined by $\displaystyle \phi_{ij}(E_{rs})=\begin{cases}1,\ (i,j)=(r,s)\\ 0, \ (i,j)\neq(r,s)\end{cases}$

    Now a symmetric matrix $\displaystyle A=(a_{ij})$ will satisfy

    $\displaystyle A=\sum_{ij}a_{ij}E_{ij}=\sum_{i}a_{ii}E_{ii}+\sum_ {i<j}a_{ij}(E_{ij}+E_{ji})$. (1)

    So a basis for $\displaystyle S$ is given by $\displaystyle S_{ii}=E_{ii}, \ S_{ij}=E_{ij}+E_{ji}$. This means we will need $\displaystyle (n^2+n)/2={\rm dim}S$ mappings for $\displaystyle S^*$. The decomposition in (1) suggests we define (for $\displaystyle 1\leq i\leq n, 1\leq i<j\leq n$) $\displaystyle \psi_{ii}:=\phi_{ii}, \ \psi_{ij}:=\phi_{ij}+\phi_{ji}$.

    Now you can prove these are linearly independent and thus form a basis. In fact, they satisfy

    $\displaystyle \psi_{ii}(S_{ii})=1, \ \psi_{ij}(S_{rs})=\begin{cases}1,\ (i,j)=(r,s)\\ 0, \ (i,j)\neq(r,s)\end{cases}$ so they form the dual basis of $\displaystyle (S_{ij})$.
    Last edited by Rebesques; Aug 31st 2007 at 08:02 PM. Reason: sleepy
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