Dual Space

**kezman** · August 31st 2007, 10:15 AM

Let $V = M_n(k)$ Vector space of n X n Matrices and $S \subset V$ Subspace of Simetric Matrices. Calculate a base for $S^o$ .

**Rebesques** · August 31st 2007, 07:57 PM

If $(E_{ij})$ are the basis matrices for $M_n$ , then you know a basis for the dual $M_n^*$ ; It consists of the $n^2$ maps

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

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

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

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

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

$\psi_{ii}(S_{ii})=1, \ \psi_{ij}(S_{rs})=\begin{cases}1,\ (i,j)=(r,s)\\ 0, \ (i,j)\neq(r,s)\end{cases}$ so they form the dual basis of $(S_{ij})$ .

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