Something more about Schur Complement

Schur Complement is a widely used way to find an inverse of a matrix in many fields. It takes place to find the decent representation of a matrix, particularly when a matrix is represented in several block sub-matrices. Assuming that an invertible matrix $M$ is ${\mathbb{R}}^{n\times n}$. A matrix $M$ can be written as four block matrices: $A$, $B$, $C$, $D$, such that $A$ and $D$ are $p\times p$ and $(n-p)\times (n-p)$ matrices respectively.

$$M=\left[\begin{array}{cc}A& B\\ C& D\end{array}\right],A\in {\mathbb{R}}^{p\times p};D\in {\mathbb{R}}^{(n-p)\times (n-p)}$$

There are many resources in the internet showing the derivation of Schur Complement. The one that I found it to be the most understandable is by Jean Gallier. He mimicked Guass elimination on block matrices to find the inverse of $M$. That is

$$\begin{gathered} M^{-1}=\left[\begin{array}{cc}(A-B{D}^{-1}C{)}^{-1}& -(A-B{D}^{-1}C{)}^{-1}B{D}^{-1}\\ -D^{-1}C(A-B{D}^{-1}C{)}^{-1}& D^{-1}+D^{-1}C(A-B{D}^{-1}C{)}^{-1}B{D}^{-1}\end{array}\right] \\ =\left[\begin{array}{cc}{A}^{-1}+A^{-1}B(D-C{A}^{-1}B{)}^{-1}C{A}^{-1}& -A^{-1}B(D-C{A}^{-1}B{)}^{-1}\\ -(D-C{A}^{-1}B{)}^{-1}C{A}^{-1}& (D-C{A}^{-1}B{)}^{-1}\end{array}\right] \end{gathered}$$

There are two ways to represent an inverse of $M$. The first way, which already shows above, takes the $D^{-1}$ to form the Schur complement. It is called Schur Complement of $D$. The latter one takes the $A^{-1}$ instead. It is worth to notice that only $A$ and $D$ are square matrices, thus constructing those two valid representations.

At this point, an inverse of a matrix M is simply constructed, but yet not the most concise one. Some books like to mention the form that further factorizes it down into three matrices.

$$M^{-1}=\left[\begin{array}{cc}{I}_p& 0\\ -D^{-1}C& I_{n-p}\end{array}\right]\left[\begin{array}{cc}(A-B{D}^{-1}C{)}^{-1}& 0\\ 0& D^{-1}\end{array}\right]\left[\begin{array}{cc}{I}_p& B{D}^{-1}\\ 0& I_{n-p}\end{array}\right]$$

To come up with the representation above, we can make use of LU decomposition to find the lower triangular matrix. We first perform linear transformation by adding $R_2$ with $D^{-1}C{R}_1$

$$\begin{gathered} \left[\begin{array}{cc}(A-B{D}^{-1}C{)}^{-1}& -(A-B{D}^{-1}C{)}^{-1}B{D}^{-1}\\ -D^{-1}C(A-B{D}^{-1}C{)}^{-1}& D^{-1}+D^{-1}C(A-B{D}^{-1}C{)}^{-1}B{D}^{-1}\end{array}\right] \\ \stackrel{R_2-(-D^{-1}C)R_1}{\to }\left[\begin{array}{cc}(A-B{D}^{-1}C{)}^{-1}& -(A-B{D}^{-1}C{)}^{-1}B{D}^{-1}\\ 0& D^{-1}\end{array}\right] \end{gathered}$$

From which, we can deduce that $K_1$, $K_2$ and $K_3$ are just $I_p$, $I_{n-p}$ and $-D^{-1}C$ respectively.

$$\begin{gathered} \left[\begin{array}{cc}{K}_1& 0\\ K_2& K_3\end{array}\right]\left[\begin{array}{cc}(A-B{D}^{-1}C{)}^{-1}& -(A-B{D}^{-1}C{)}^{-1}B{D}^{-1}\\ 0& D^{-1}\end{array}\right]=M^{-1} \\ \left[\begin{array}{cc}{K}_1& 0\\ K_2& K_3\end{array}\right]=\left[\begin{array}{cc}{I}_p& 0\\ -D^{-1}C& I_{n-p}\end{array}\right] \end{gathered}$$

We then undergo the similar step on the matrix to find its upper triangular matrix by performing linear transaction one more time. This time, we add $R_1$ by $(A-B{D}^{-1}C{)}^{-1}{R}_2$.

$$\left[\begin{array}{cc}(A-B{D}^{-1}C{)}^{-1}& -(A-B{D}^{-1}C{)}^{-1}B{D}^{-1}\\ 0& D^{-1}\end{array}\right]\stackrel{R_1+(A-B{D}^{-1}C{)}^{-1}B{R}_2}{\to }\left[\begin{array}{cc}(A-B{D}^{-1}C{)}^{-1}& 0\\ 0& D^{-1}\end{array}\right]$$

Again, we can deduce that $U_1$, $U_2$ and $U_3$ are $I_p$, $-B{D}^{-1}$ and $I_{n-p}$.

$$\begin{gathered} \left[\begin{array}{cc}(A-B{D}^{-1}C{)}^{-1}& 0\\ 0& D^{-1}\end{array}\right]\left[\begin{array}{cc}{U}_1& U_2\\ 0& U_3\end{array}\right]=\left[\begin{array}{cc}(A-B{D}^{-1}C{)}^{-1}& -(A-B{D}^{-1}C{)}^{-1}B{D}^{-1}\\ 0& D^{-1}\end{array}\right] \\ \left[\begin{array}{cc}{U}_1& U_2\\ 0& U_3\end{array}\right]=\left[\begin{array}{cc}{I}_p& -B{D}^{-1}\\ 0& I_{n-p}\end{array}\right] \end{gathered}$$

After all, we factorize the Schur complement of an inverse $M$ into a more concise form.

$$M^{-1}=\left[\begin{array}{cc}{I}_p& 0\\ -D^{-1}C& I_{n-p}\end{array}\right]\left[\begin{array}{cc}(A-B{D}^{-1}C{)}^{-1}& 0\\ 0& D^{-1}\end{array}\right]\left[\begin{array}{cc}{I}_p& -B{D}^{-1}\\ 0& I_{n-p}\end{array}\right]$$

References:

• The Schur Complement and Symmetric Positive Semidefinite (and Definite) Matrices
• LU Decomposition Wikipedia