The matrix
S
is symmetric, since
S^{T} | = | D^{-1}_{n}V_{n}P^{T}_{n}J_{n}^{-1}J^{-T}_{n}P_{n} | |
= | D^{-1}_{n}V_{n}P^{T}_{n}J_{n}^{-1}J^{-T}_{n}P_{n} | ||
= | D^{-1}_{n}V_{n}J^{-T}_{n}P_{n}P^{T}_{n}J_{n}^{-1} | ||
= | D^{-1}_{n}V_{n}P_{n}^{T}J_{n}^{-1}J^{-T}_{n}P_{n} | ||
= | S |
x^{T}Sx | = | D^{-1}_{n}V_{n}x^{T}P^{T}_{n}J_{n}^{-1}J^{-T}_{n}P_{n}x | |
= | D^{-1}_{n}V_{n}J^{-T}_{n}P_{n}xJ^{-T}_{n}P_{n}x | ||
= | D^{-1}_{n}V_{n}J^{-T}_{n}P_{n}x | ||
> | 0 |
This is necessary, but not sufficient, to proving that the entire method is SPD. See the associated paper for the gory details.