Editing Schur complement (section)

==Background==
The Schur complement arises when performing a block [[Gaussian elimination]] on the matrix ''M''. In order to eliminate the elements below the block diagonal, one multiplies the matrix ''M'' by a ''block lower triangular'' matrix on the right as follows:
<math display="block">\begin{align}
   &M = \begin{bmatrix} A & B \\ C & D \end{bmatrix} \quad \to \quad \begin{bmatrix} A & B \\ C & D \end{bmatrix} \begin{bmatrix} I_p & 0 \\ -D^{-1}C & I_q \end{bmatrix} = \begin{bmatrix} A - BD^{-1}C & B \\ 0 & D \end{bmatrix},
\end{align}</math>
where ''I<sub>p</sub>'' denotes a ''p''&times;''p'' [[identity matrix]]. As a result, the Schur complement <math>M/D = A - BD^{-1}C</math> appears in the upper-left ''p''&times;''p'' block.

Continuing the elimination process beyond this point (i.e., performing a block [[Gauss–Jordan elimination]]),
<math display="block">\begin{align}
     &\begin{bmatrix} A - BD^{-1}C & B \\ 0 & D \end{bmatrix} \quad \to \quad  \begin{bmatrix} I_p & -BD^{-1} \\ 0 & I_q \end{bmatrix} \begin{bmatrix} A - BD^{-1}C & B \\ 0 & D \end{bmatrix} 
= \begin{bmatrix} A - BD^{-1}C & 0 \\ 0 & D \end{bmatrix},
\end{align}</math>
leads to an [[LDU decomposition]] of ''M'', which reads
<math display="block">\begin{align}
  M &= \begin{bmatrix} A & B \\ C & D \end{bmatrix} 
      = \begin{bmatrix} I_p & BD^{-1} \\ 0 & I_q \end{bmatrix}\begin{bmatrix} A - BD^{-1}C & 0 \\ 0 & D \end{bmatrix}\begin{bmatrix} I_p & 0 \\ D^{-1}C & I_q \end{bmatrix}.
\end{align}</math>
Thus, the inverse of ''M'' may be expressed involving ''D''<sup>&minus;1</sup> and the inverse of Schur's complement, assuming it exists, as
<math display="block">\begin{align}
      M^{-1} = \begin{bmatrix} A & B \\ C & D \end{bmatrix}^{-1}
  ={} &\left(\begin{bmatrix} I_p & BD^{-1} \\ 0 & I_q \end{bmatrix}
         \begin{bmatrix} A - BD^{-1}C & 0 \\ 0 & D \end{bmatrix}
         \begin{bmatrix} I_p & 0 \\ D^{-1}C & I_q \end{bmatrix}
       \right)^{-1} \\
  ={} &\begin{bmatrix}                            I_p & 0 \\ -D^{-1}C & I_q    \end{bmatrix}
         \begin{bmatrix} \left(A - BD^{-1}C\right)^{-1} & 0 \\        0 & D^{-1} \end{bmatrix}
         \begin{bmatrix} I_p & -BD^{-1} \\ 0 & I_q \end{bmatrix} \\[4pt]
  ={} &\begin{bmatrix}
          \left(A - BD^{-1}C\right)^{-1}        & -\left(A - BD^{-1}C\right)^{-1} BD^{-1} \\
         -D^{-1}C\left(A - BD^{-1}C\right)^{-1} &  D^{-1} + D^{-1}C\left(A - BD^{-1}C\right)^{-1}BD^{-1}
       \end{bmatrix} \\[4pt]
  ={} &\begin{bmatrix}
          \left(M/D\right)^{-1}        & -\left(M/D\right)^{-1} B D^{-1} \\
         -D^{-1}C\left(M/D\right)^{-1} &  D^{-1} + D^{-1}C\left(M/D\right)^{-1} B D^{-1}
       \end{bmatrix}.
\end{align}</math>
The above relationship comes from the elimination operations that involve ''D''<sup>&minus;1</sup> and ''M/D''. An equivalent derivation can be done with the roles of ''A'' and ''D'' interchanged. By equating the expressions for ''M''<sup>&minus;1</sup> obtained in these two different ways, one can establish the [[matrix inversion lemma]], which relates the two Schur complements of ''M'': ''M/D'' and ''M/A'' (see ''"Derivation from LDU decomposition"'' in {{slink|Woodbury_matrix_identity#Alternative_proofs}}).