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Support Operator Method Derivation: SPD Proof

The matrix S is symmetric, since

ST = $\displaystyle \left[\vphantom{ \sum_n D^{-1}_n V_n {\mathbf{P}}^{\mathrm{T}}_n
\mathbf{J}_n^{-1} {\mathbf{J}}^{-\mathrm{T}}_n
\mathbf{P}_n }\right.$$\displaystyle \sum_{n}^{}$D-1nVnPTnJn-1J-TnPn$\displaystyle \left.\vphantom{ \sum_n D^{-1}_n V_n {\mathbf{P}}^{\mathrm{T}}_n
...
...n^{-1} {\mathbf{J}}^{-\mathrm{T}}_n
\mathbf{P}_n }\right]^{{{\mathrm{T}}}}_{{}}$  
  = $\displaystyle \sum_{n}^{}$D-1nVn$\displaystyle \left[\vphantom{ {\mathbf{P}}^{\mathrm{T}}_n
\mathbf{J}_n^{-1} {\mathbf{J}}^{-\mathrm{T}}_n
\mathbf{P}_n }\right.$PTnJn-1J-TnPn$\displaystyle \left.\vphantom{ {\mathbf{P}}^{\mathrm{T}}_n
\mathbf{J}_n^{-1} {\mathbf{J}}^{-\mathrm{T}}_n
\mathbf{P}_n }\right]^{{{\mathrm{T}}}}_{{}}$  
  = $\displaystyle \sum_{n}^{}$D-1nVn$\displaystyle \left[\vphantom{ {\mathbf{J}}^{-\mathrm{T}}_n
\mathbf{P}_n }\right.$J-TnPn$\displaystyle \left.\vphantom{ {\mathbf{J}}^{-\mathrm{T}}_n
\mathbf{P}_n }\right]^{{{\mathrm{T}}}}_{{}}$$\displaystyle \left[\vphantom{ {\mathbf{P}}^{\mathrm{T}}_n
\mathbf{J}_n^{-1} }\right.$PTnJn-1$\displaystyle \left.\vphantom{ {\mathbf{P}}^{\mathrm{T}}_n
\mathbf{J}_n^{-1} }\right]^{{{\mathrm{T}}}}_{{}}$  
  = $\displaystyle \sum_{n}^{}$D-1nVnPnTJn-1J-TnPn  
  = S  

The matrix S is positive definite, since
xTSx = $\displaystyle \sum_{n}^{}$D-1nVnxTPTnJn-1J-TnPnx  
  = $\displaystyle \sum_{n}^{}$D-1nVn$\displaystyle \left[\vphantom{
{\mathbf{J}}^{-\mathrm{T}}_n \mathbf{P}_n \mathbf{x} }\right.$J-TnPnx$\displaystyle \left.\vphantom{
{\mathbf{J}}^{-\mathrm{T}}_n \mathbf{P}_n \mathbf{x} }\right]^{{{\mathrm{T}}}}_{{}}$$\displaystyle \left[\vphantom{
{\mathbf{J}}^{-\mathrm{T}}_n \mathbf{P}_n \mathbf{x} }\right.$J-TnPnx$\displaystyle \left.\vphantom{ {\mathbf{J}}^{-\mathrm{T}}_n \mathbf{P}_n \mathbf{x} }\right]$  
  = $\displaystyle \sum_{n}^{}$D-1nVn$\displaystyle \left\Vert\vphantom{
{\mathbf{J}}^{-\mathrm{T}}_n \mathbf{P}_n \mathbf{x} }\right.$J-TnPnx$\displaystyle \left.\vphantom{
{\mathbf{J}}^{-\mathrm{T}}_n \mathbf{P}_n \mathbf{x} }\right\Vert _{2}^{}$  
  > 0$\displaystyle \mbox{\hspace{2em} if $D^{-1}_n V_n > 0$\ and
${\mathbf{J}}^{-\mathrm{T}}_n \mathbf{P}_n \mathbf{x} \neq 0$}$  

If S is SPD, then S-1 is also symmetric positive definite.

This is necessary, but not sufficient, to proving that the entire method is SPD. See the associated paper for the gory details.


next up previous
Next: Second-Order Demonstration Up: Hexahedral Support Operator Method Previous: Support Operator Method Derivation:
Michael L. Hall