Support Operator Method Derivation: SPD Proof

The matrix S is symmetric, since

$$\left[\vphantom{ \sum_n D^{-1}_n V_n {\mathbf{P}}^{\mathrm{T}}_n \mathbf{J}_n^{-1} {\mathbf{J}}^{-\mathrm{T}}_n \mathbf{P}_n }\right. \sum_{n}^{} \left.\vphantom{ \sum_n D^{-1}_n V_n {\mathbf{P}}^{\mathrm{T}}_n ... ...n^{-1} {\mathbf{J}}^{-\mathrm{T}}_n \mathbf{P}_n }\right]^{{{\mathrm{T}}}}_{{}}$$ S^T =

$$\sum_{n}^{} \left[\vphantom{ {\mathbf{P}}^{\mathrm{T}}_n \mathbf{J}_n^{-1} {\mathbf{J}}^{-\mathrm{T}}_n \mathbf{P}_n }\right. \left.\vphantom{ {\mathbf{P}}^{\mathrm{T}}_n \mathbf{J}_n^{-1} {\mathbf{J}}^{-\mathrm{T}}_n \mathbf{P}_n }\right]^{{{\mathrm{T}}}}_{{}}$$ =

$$\sum_{n}^{} \left[\vphantom{ {\mathbf{J}}^{-\mathrm{T}}_n \mathbf{P}_n }\right. \left.\vphantom{ {\mathbf{J}}^{-\mathrm{T}}_n \mathbf{P}_n }\right]^{{{\mathrm{T}}}}_{{}} \left[\vphantom{ {\mathbf{P}}^{\mathrm{T}}_n \mathbf{J}_n^{-1} }\right. \left.\vphantom{ {\mathbf{P}}^{\mathrm{T}}_n \mathbf{J}_n^{-1} }\right]^{{{\mathrm{T}}}}_{{}}$$ =

$$\sum_{n}^{}$$ =

= S

The matrix S is positive definite, since

$$\sum_{n}^{}$$ x^TSx =

$$\sum_{n}^{} \left[\vphantom{ {\mathbf{J}}^{-\mathrm{T}}_n \mathbf{P}_n \mathbf{x} }\right. \left.\vphantom{ {\mathbf{J}}^{-\mathrm{T}}_n \mathbf{P}_n \mathbf{x} }\right]^{{{\mathrm{T}}}}_{{}} \left[\vphantom{ {\mathbf{J}}^{-\mathrm{T}}_n \mathbf{P}_n \mathbf{x} }\right. \left.\vphantom{ {\mathbf{J}}^{-\mathrm{T}}_n \mathbf{P}_n \mathbf{x} }\right]$$ =

$$\sum_{n}^{} \left\Vert\vphantom{ {\mathbf{J}}^{-\mathrm{T}}_n \mathbf{P}_n \mathbf{x} }\right. \left.\vphantom{ {\mathbf{J}}^{-\mathrm{T}}_n \mathbf{P}_n \mathbf{x} }\right\Vert _{2}^{}$$ =

$$\mbox{\hspace{2em} if $D^{-1}_n V_n > 0$\ and ${\mathbf{J}}^{-\mathrm{T}}_n \mathbf{P}_n \mathbf{x} \neq 0$}$$ >

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.