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Previous: Support Operator Method Derivation:
The matrix
S
is symmetric, since
ST |
= |
![$\displaystyle \left[\vphantom{ \sum_n D^{-1}_n V_n {\mathbf{P}}^{\mathrm{T}}_n
\mathbf{J}_n^{-1} {\mathbf{J}}^{-\mathrm{T}}_n
\mathbf{P}_n }\right.$](img106.gif) D-1nVnPTnJn-1J-TnPn![$\displaystyle \left.\vphantom{ \sum_n D^{-1}_n V_n {\mathbf{P}}^{\mathrm{T}}_n
...
...n^{-1} {\mathbf{J}}^{-\mathrm{T}}_n
\mathbf{P}_n }\right]^{{{\mathrm{T}}}}_{{}}$](img108.gif) |
|
|
= |
D-1nVn PTnJn-1J-TnPn![$\displaystyle \left.\vphantom{ {\mathbf{P}}^{\mathrm{T}}_n
\mathbf{J}_n^{-1} {\mathbf{J}}^{-\mathrm{T}}_n
\mathbf{P}_n }\right]^{{{\mathrm{T}}}}_{{}}$](img110.gif) |
|
|
= |
D-1nVn J-TnPn![$\displaystyle \left.\vphantom{ {\mathbf{J}}^{-\mathrm{T}}_n
\mathbf{P}_n }\right]^{{{\mathrm{T}}}}_{{}}$](img112.gif) PTnJn-1![$\displaystyle \left.\vphantom{ {\mathbf{P}}^{\mathrm{T}}_n
\mathbf{J}_n^{-1} }\right]^{{{\mathrm{T}}}}_{{}}$](img114.gif) |
|
|
= |
D-1nVnPnTJn-1J-TnPn |
|
|
= |
S |
|
The matrix
S
is positive definite, since
xTSx |
= |
D-1nVnxTPTnJn-1J-TnPnx |
|
|
= |
D-1nVn J-TnPnx![$\displaystyle \left.\vphantom{
{\mathbf{J}}^{-\mathrm{T}}_n \mathbf{P}_n \mathbf{x} }\right]^{{{\mathrm{T}}}}_{{}}$](img116.gif) J-TnPnx![$\displaystyle \left.\vphantom{ {\mathbf{J}}^{-\mathrm{T}}_n \mathbf{P}_n \mathbf{x} }\right]$](img117.gif) |
|
|
= |
D-1nVn J-TnPnx![$\displaystyle \left.\vphantom{
{\mathbf{J}}^{-\mathrm{T}}_n \mathbf{P}_n \mathbf{x} }\right\Vert _{2}^{}$](img119.gif) |
|
|
> |
0![$\displaystyle \mbox{\hspace{2em} if $D^{-1}_n V_n > 0$\ and
${\mathbf{J}}^{-\mathrm{T}}_n \mathbf{P}_n \mathbf{x} \neq 0$}$](img120.gif) |
|
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: Second-Order Demonstration
Up: Hexahedral Support Operator Method
Previous: Support Operator Method Derivation:
Michael L. Hall