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Schur complement

by , PhD

The Schur complements of a block matrix are functions of its blocks that allow to derive several useful formulae for the inversion and the factorization of the block matrix itself.

Table of Contents

Definition

Let us start with a definition.

Definition Let $M$ be a block matrix[eq1]such that its blocks A and $B$ are square matrices. If A is invertible, the Schur complement of A in $M$ is[eq2]If $D$ is invertible, the Schur complement of $D$ in $M$ is[eq3]

Note that the definition does not specify the dimensions of $M$ and its blocks. The only essential thing that we need to specify is that A and $B$ must be square matrices. Everything else follows from the latter requirement.

In fact, suppose that A is $K_{A}	imes K_{A}$ and $D$ is $K_{D}	imes K_{D}$. Then, $B$ must be $K_{A}	imes K_{D}$ and $C$ must be $K_{D}	imes K_{A}$. As a consequence, the product $CA^{-1}B$ is well-defined and its dimension is $K_{D}	imes K_{D}$. Similarly, the product $BD^{-1}C$ is well-defined and its dimension is $K_{A}	imes K_{A}$.

Role in block matrix inversion

Schur complements play a key role in the inversion of block matrices.

Proposition Let $M$ be a block matrix[eq4]such that its blocks A and $B$ are square matrices. If A and its Schur complement $M/A$ are invertible, then $M$ is invertible and[eq5]

Proof

The rigorous way to prove the proposition is to multiply $M$ by $M^{-1}$ using the rule for the multiplication of block matrices and to show that the result is the identity matrix. However, we take an approach that shows how the formulae for the blocks of $M^{-1}$ have been derived. Suppose that $M^{-1}$ exists and partition it into blocks as follows:[eq6]where $lpha $ has the same dimension as A, and $D$ has the same dimension as $delta $. By the definition of inverse, $M^{-1}$ should satisfy[eq7]or[eq8]where $I_{A}$ is the identity matrix having the same dimension as A, and $I_{D}$ is the identity matrix having the same dimension as $D$. This is equivalent to saying that the four blocks [eq9] should satisfy the four equations[eq10]If A is invertible, we can transform the first equation into[eq11]Then we substitute equation (5) into (3) and get[eq12]or[eq13]Thus, the Schur complement of A has already appeared. Let us write it explicitly:[eq14]If the Schur complement of A is invertible, then we have[eq15]Thus, we have found the $left( 2,1
ight) $-th block of $M^{-1}$. We can now plug equation (6) into (5) and obtain another block of $M^{-1}$:[eq16]The next step is to transform equation (2) as follows:[eq17]Then, we plug equation (7) into (4):[eq18]From the latter equation, we obtain one more block of $M^{-1}$:[eq19]We can now substitute equation (8) into (7) and recover the last block of $M^{-1}$:[eq20]

There is an analogous proposition for the Schur complement of $D$. You may try to prove it as an exercise and then use the proof below to check your solution.

Proposition Let $M$ be a block matrix[eq21]such that its blocks A and $B$ are square matrices. If $D$ and its Schur complement $D/A$ are invertible, then $M$ is invertible and[eq22]

Proof

As before, suppose that $M^{-1}$ exists and partition it into blocks:[eq23]where $lpha $ has the same dimension as A, and $D$ has the same dimension as $delta $. By the definition of inverse, $M^{-1}$ should satisfy[eq7]or[eq8]where $I_{A}$ is the identity matrix having the same dimension as A, and $I_{D}$ is the identity matrix having the same dimension as $D$. Thus, the four blocks [eq26] need to satisfy the equations[eq10]Since $D$ is invertible, then equation (4) can be written as[eq28]By plugging equation (5) into (2), we obtain[eq29]or[eq30]By using the appropriate notation for the Schur complement of $D$, we can write[eq31]Since the Schur complement of $D$ is invertible, we can solve the latter equation and get our first solution for a block of $M^{-1}$:[eq32]Then we substitute equation (6) into (5) and obtain another block of $M^{-1}$:[eq33]We now manipulate equation (3):[eq34]and substitute (7) into (1):[eq35]Thus, we derive one more block of $M^{-1}$:[eq36]Finally, the last block of $M^{-1}$ is obtained by plugging equation (8) into (7):[eq37]

Factorization of a block matrix

The Schur complements are often used to factorize a block matrix into a product of simpler block matrices.

Proposition Let $M$ be a block matrix[eq38]such that its blocks A and $B$ are square matrices. If A is invertible, then[eq39]where I are identity matrices and 0 are matrices of zeros.

Proof

Remember that the Schur complement of A in $M$ is[eq40]Therefore, the product of the three matrices is[eq41]

Proposition Let $M$ be a block matrix[eq38]such that its blocks A and $B$ are square matrices. If $D$ is invertible, then[eq43]where I are identity matrices and 0 are matrices of zeros.

Proof

Remember that the Schur complement of $D$ in $M$ is[eq44]Therefore, the product of the three matrices is[eq45]

Factorization of the inverse of a block matrix

When the Schur complements are invertible, they can be used to derive useful factorizations of the inverse of a block matrix.

Proposition Let $M$ be a block matrix[eq38]such that its blocks A and $B$ are square matrices. If A and its Schur complement are invertible, then[eq47]where I are identity matrices and 0 are matrices of zeros.

Proof

If we multiply the factorization of $M$ into three matrices derived above by the factorization of $M^{-1}$ proposed here, we obtain the identity matrix because[eq48]

Proposition Let $M$ be a block matrix[eq38]such that its blocks A and $B$ are square matrices. If $D$ and its Schur complement are invertible, then[eq50]where I are identity matrices and 0 are matrices of zeros.

Proof

If we multiply the factorization of $M$ into three matrices derived above by the factorization of $M^{-1}$ proposed here, we obtain the identity matrix because[eq51]

Solved exercises

Below you can find some exercises with explained solutions.

Exercise 1

Let $M$ be a block matrix[eq52]

Suppose that A and $D$ are invertible and $B$ and $C$ are zero. Invert $M$ by using the Schur complement of A.

Solution

The Schur complement of A is[eq53]and its inverse is[eq54]Therefore,[eq55]

How to cite

Please cite as:

Taboga, Marco (2017). "Schur complement", Lectures on matrix algebra. https://www.statlect.com/matrix-algebra/Schur-complement.

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