Schur Complement

The Schur complement is defined for a block matrix $M$ that is partitioned into four sub-blocks:

$$M=\left[\begin{array}{cc}A& B\\ C& D\end{array}\right]$$

where $A$ and $D$ are square matrices.

Schur Complement of $\mathbf{D}$

If the block $D$ is invertible, the Schur complement of $D$ in $M$ is the matrix $M/D$ defined as:

$$M/D=A-B{D}^{-1}C$$

Schur Complement of $\mathbf{A}$

Similarly, if the block $A$ is invertible, the Schur complement of $A$ in $M$ is the matrix $M/A$ defined as:

$$M/A=D-C{A}^{-1}B$$

Decomposition

The cool thing is that if the complement is possible, then we can decompose $\mathbf{M}$

Decomposition via Schur Complement of $\mathbf{A}$

This decomposition is possible if the top-left block $\mathbf{A}$ is invertible. The Schur Complement of $A$ in $M$ is:

$$M/A=D-C{A}^{-1}B$$

The matrix $M$ can then be decomposed as:

$$M=\left[\begin{array}{cc}I& 0\\ C{A}^{-1}& I\end{array}\right]\left[\begin{array}{cc}A& 0\\ 0& M/A\end{array}\right]\left[\begin{array}{cc}I& A^{-1}B\\ 0& I\end{array}\right]$$

Where:

• The first matrix is block lower triangular ($L$).
• The second matrix is block diagonal ($D$) and contains the original block $A$ and the Schur complement $M/A$.
• The third matrix is block upper triangular ($U$).

This formula shows that $M$ is congruent to the block-diagonal matrix $\mathrm{diag}(A,M/A)$ via a block triangular matrix.

Decomposition via Schur Complement of $\mathbf{D}$

This decomposition is possible if the bottom-right block $\mathbf{D}$ is invertible. The Schur Complement of $D$ in $M$ is:

$$M/D=A-B{D}^{-1}C$$

The matrix $M$ can then be decomposed as:

$$M=\left[\begin{array}{cc}I& B{D}^{-1}\\ 0& I\end{array}\right]\left[\begin{array}{cc}M/D& 0\\ 0& D\end{array}\right]\left[\begin{array}{cc}I& 0\\ D^{-1}C& I\end{array}\right]$$

Where:

• The first matrix is block upper triangular ($U^{\prime }$).
• The second matrix is block diagonal ($D^{\prime }$) and contains the Schur complement $M/D$ and the original block $D$.
• The third matrix is block lower triangular ($L^{\prime }$).