# Schur complement

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In linear algebra and the theory of matrices, the Schur complement of a block matrix is defined as follows.

Suppose A, B, C, D are respectively p × p, p × q, q × p, and q × q matrices, and D is invertible. Let

$M=\left[{\begin{matrix}A&B\\C&D\end{matrix}}\right]$ so that M is a (p + q) × (p + q) matrix.

Then the Schur complement of the block D of the matrix M is the p × p matrix defined by

$M/D:=A-BD^{-1}C\,$ and, if A is invertible, the Schur complement of the block A of the matrix M is the q × q matrix defined by

$M/A:=D-CA^{-1}B.$ In the case that A or D is singular, substituting a generalized inverse for the inverses on M/A and M/D yields the generalized Schur complement.

The Schur complement is named after Issai Schur who used it to prove Schur's lemma, although it had been used previously. Emilie Haynsworth was the first to call it the Schur complement. The Schur complement is a key tool in the fields of numerical analysis, statistics and matrix analysis.

## Background

The Schur complement arises as the result of performing a block Gaussian elimination by multiplying the matrix M from the right with a block lower triangular matrix

$L={\begin{bmatrix}I_{p}&0\\-D^{-1}C&I_{q}\end{bmatrix}}.$ Here Ip denotes a p×p identity matrix. After multiplication with the matrix L the Schur complement appears in the upper p×p block. The product matrix is

{\begin{aligned}ML&={\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}}\\[4pt]&={\begin{bmatrix}I_{p}&BD^{-1}\\0&I_{q}\end{bmatrix}}{\begin{bmatrix}A-BD^{-1}C&0\\0&D\end{bmatrix}}.\end{aligned}} This is analogous to an LDU decomposition. That is, we have shown that

{\begin{aligned}{\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{aligned}} and inverse of M thus may be expressed involving D−1 and the inverse of Schur's complement (if it exists) only as

{\begin{aligned}&{\begin{bmatrix}A&B\\C&D\end{bmatrix}}^{-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(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}&\left(D-CA^{-1}B\right)^{-1}\end{bmatrix}}\\[4pt]={}&{\begin{bmatrix}\left(M/D\right)^{-1}&-\left(M/D\right)^{-1}BD^{-1}\\-D^{-1}C\left(M/D\right)^{-1}&\left(M/A\right)^{-1}\end{bmatrix}}.\end{aligned}} Cf. matrix inversion lemma which illustrates relationships between the above and the equivalent derivation with the roles of A and D interchanged.

## Properties

• If M is a positive-definite symmetric matrix, then so is the Schur complement of D in M.
• If p and q are both 1 (i.e., A, B, C and D are all scalars), we get the familiar formula for the inverse of a 2-by-2 matrix:
$M^{-1}={\frac {1}{AD-BC}}\left[{\begin{matrix}D&-B\\-C&A\end{matrix}}\right]$ provided that AD − BC is non-zero.
• In general, if A is invertible, then
$M^{-1}={\begin{bmatrix}A^{-1}+A^{-1}B(M/A)^{-1}CA^{-1}&-A^{-1}B(M/A)^{-1}\\-(M/A)^{-1}CA^{-1}&(M/A)^{-1}\end{bmatrix}}$ whenever this inverse exists.
• The determinant of M is also clearly seen to be given by
$\det(M)=\det(D)\det \left(A-BD^{-1}C\right)$ which generalizes the determinant formula for 2 × 2 matrices.
• (Guttman rank additivity formula) The rank of M is given by
$\operatorname {rank} (M)=\operatorname {rank} (D)+\operatorname {rank} \left(A-BD^{-1}C\right)$ • (Haynsworth inertia additivity formula) The inertia of the block matrix M is equal to the inertia of A plus the inertia of M/A.

## Application to solving linear equations

The Schur complement arises naturally in solving a system of linear equations such as

{\begin{aligned}Ax+By&=a\\Cx+Dy&=b\end{aligned}} where x, a are p-dimensional column vectors, y, b are q-dimensional column vectors, and A, B, C, D are as above. Multiplying the bottom equation by ${\textstyle BD^{-1}}$ and then subtracting from the top equation one obtains

$\left(A-BD^{-1}C\right)x=a-BD^{-1}b.$ Thus if one can invert D as well as the Schur complement of D, one can solve for x, and then by using the equation ${\textstyle Cx+Dy=b}$ one can solve for y. This reduces the problem of inverting a ${\textstyle (p+q)\times (p+q)}$ matrix to that of inverting a p × p matrix and a q × q matrix. In practice, one needs D to be well-conditioned in order for this algorithm to be numerically accurate.

In electrical engineering this is often referred to as node elimination or Kron reduction.

## Applications to probability theory and statistics

Suppose the random column vectors X, Y live in Rn and Rm respectively, and the vector (X, Y) in Rn + m has a multivariate normal distribution whose covariance is the symmetric positive-definite matrix

$\Sigma =\left[{\begin{matrix}A&B\\B^{\mathsf {T}}&C\end{matrix}}\right],$ where ${\textstyle A\in \mathbb {R} ^{n\times n}}$ is the covariance matrix of X, ${\textstyle C\in \mathbb {R} ^{m\times m}}$ is the covariance matrix of Y and ${\textstyle B\in \mathbb {R} ^{n\times m}}$ is the covariance matrix between X and Y.

Then the conditional covariance of X given Y is the Schur complement of C in ${\textstyle \Sigma }$ :

{\begin{aligned}\operatorname {Cov} (X\mid Y)&=A-BC^{-1}B^{\mathsf {T}}\\\operatorname {E} (X\mid Y)&=\operatorname {E} (X)+BC^{-1}(Y-\operatorname {E} (Y))\end{aligned}} If we take the matrix $\Sigma$ above to be, not a covariance of a random vector, but a sample covariance, then it may have a Wishart distribution. In that case, the Schur complement of C in $\Sigma$ also has a Wishart distribution.[citation needed]

## Schur complement condition for positive definiteness and positive semi-definiteness

Let X be a symmetric matrix given by

$X=\left[{\begin{matrix}A&B\\B^{\mathsf {T}}&C\end{matrix}}\right].$ Let X/A be the Schur complement of A in X; i.e.,

$X/A=C-B^{\mathsf {T}}A^{-1}B,\,$ and X/C be the Schur complement of C in X; i.e.,

$X/C=A-BC^{-1}B^{\mathsf {T}}.\,$ Then

• X is positive definite if and only if A and X/A are both positive definite:
$X\succ 0\Leftrightarrow A\succ 0,X/A=C-B^{\mathsf {T}}A^{-1}B\succ 0.$ • X is positive definite if and only if C and X/C are both positive definite:
$X\succ 0\Leftrightarrow C\succ 0,X/C=A-BC^{-1}B^{\mathsf {T}}\succ 0.$ • If A is positive definite, then X is positive semi-definite if and only if X/A is positive semi-definite:
${\text{If }}A\succ 0,{\text{ then }}X\succeq 0\Leftrightarrow X/A=C-B^{\mathsf {T}}A^{-1}B\succeq 0.$ • If C is positive definite, then X is positive semi-definite if and only if X/C is positive semi-definite:
${\text{If }}C\succ 0,{\text{ then }}X\succeq 0\Leftrightarrow X/C=A-BC^{-1}B^{\mathsf {T}}\succeq 0.$ The first and third statements can be derived by considering the minimizer of the quantity

$u^{\mathsf {T}}Au+2v^{\mathsf {T}}B^{\mathsf {T}}u+v^{\mathsf {T}}Cv,\,$ as a function of v (for fixed u).

Furthermore, since

$\left[{\begin{matrix}A&B\\B^{\mathsf {T}}&C\end{matrix}}\right]\succ 0\Longleftrightarrow \left[{\begin{matrix}C&B^{\mathsf {T}}\\B&A\end{matrix}}\right]\succ 0$ and similarly for positive semi-definite matrices, the second (respectively fourth) statement is immediate from the first (resp. third) statement.

There is also a sufficient and necessary condition for the positive semi-definiteness of X in terms of a generalized Schur complement. Precisely,

• $X\succeq 0\Leftrightarrow A\succeq 0,C-B^{\mathsf {T}}A^{g}B\succeq 0,\left(I-AA^{g}\right)B=0\,$ and
• $X\succeq 0\Leftrightarrow C\succeq 0,A-BC^{g}B^{\mathsf {T}}\succeq 0,\left(I-CC^{g}\right)B^{\mathsf {T}}=0,$ where $A^{g}$ denotes the generalized inverse of $A$ .