# Schur complement in LUP decomposition of a block tridiagonal matrix

Section 2.2 of the article On twisted factorizations of block tridiagonal matrices explains how to do a LUP decomposition of block tridiagonal matrices by showing the process on a 4 blocks by 4 blocks matrix. It involves finding the LUP decomposition of a matrix that is supposedly a Schur complement, but i can't understand what it is the Schur complement of. I rewrote the important parts here :

$$W = \begin{pmatrix}B_1&C_1&&\\A_2&B_2&C_2&\\&A_3&B_3&C_3\\&&A_4&B_4\end{pmatrix}= PLU= \begin{pmatrix}P_1&&&\\&P_2&&\\&&P_3&\\&&&P_4\end{pmatrix} \begin{pmatrix}L_1&&&\\M_2&L_2&&\\&M_3&L_3&\\&&M_4&L_4\end{pmatrix} \begin{pmatrix}U_1&N_1&&\\&U_2&N_2&\\&&U_3&N_3\\&&&U_4\end{pmatrix}= \begin{pmatrix}P_1L_1U_1&P_1L_1N_1&&\\P_2M_2U_1&P_2L_2U_2+P_2M_2N_1&P_2L_2N_2&\\&P_3M_3U_2&P_3L_3U_3 + P_3M_3N_2&P_3L_3N_3\\&&P_4M_4U_3&P_4L_4U_4+P_4M_4N_3\end{pmatrix}$$ With $$P$$,$$L$$ and $$U$$ to be determined.

First they find $$P_1,L_1,U_1$$ from $$B_1 = P_1L_1U_1$$, then $$N_1$$ and $$M'_2:=P_2M_2$$ from $$P_1L_1N_1 = C_1$$ and $$P_2M_2U_1 = A_2$$. Afterward is the part that i'm looking to be cleared up. They get $$P_2,L_2,U_2$$ from $$B_2=P_2L_2U_2 + M'_2$$ rearanged into $$B_2 - M'_2N_1 = P_2L_2U_2$$. This is done, in their words, by "factorizing the Schur complement $$B_2 - M'_2N_1$$ with partial pivoting".

I don't get what $$B_2 - M'_2N_1$$ is the Schur complement of, so i was hoping someone could help clear that up.