Schur complement

In linear algebra and the theory of matrices, the Schur complement of a matrix block (i.e., a submatrix within a larger 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

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

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

${\displaystyle M/D=A-BD^{-1}C\,}$

and the Schur complement of the block A of the matrix M is the q×q matrix

${\displaystyle M/A=D-CA^{-1}B.\,}$

In the case that A or D is singular, the inverses on M/A and M/D can be replaced by a generalized inverse, yielding what is called 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.[1] Emilie Haynsworth was the first to call it the Schur complement.[2] 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 the "block lower triangular" matrix

${\displaystyle L=\left[{\begin{matrix}I_{p}&0\\-D^{-1}C&I_{q}\end{matrix}}\right].}$

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

{\displaystyle {\begin{aligned}ML&=\left[{\begin{matrix}A&B\\C&D\end{matrix}}\right]\left[{\begin{matrix}I_{p}&0\\-D^{-1}C&I_{q}\end{matrix}}\right]=\left[{\begin{matrix}A-BD^{-1}C&B\\0&D\end{matrix}}\right]\\&=\left[{\begin{matrix}I_{p}&BD^{-1}\\0&I_{q}\end{matrix}}\right]\left[{\begin{matrix}A-BD^{-1}C&0\\0&D\end{matrix}}\right].\end{aligned}}}

This is analogous to an LDU decomposition. That is, we have shown that

{\displaystyle {\begin{aligned}\left[{\begin{matrix}A&B\\C&D\end{matrix}}\right]&=\left[{\begin{matrix}I_{p}&BD^{-1}\\0&I_{q}\end{matrix}}\right]\left[{\begin{matrix}A-BD^{-1}C&0\\0&D\end{matrix}}\right]\left[{\begin{matrix}I_{p}&0\\D^{-1}C&I_{q}\end{matrix}}\right],\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

{\displaystyle {\begin{aligned}&{}\quad \left[{\begin{matrix}A&B\\C&D\end{matrix}}\right]^{-1}=\left[{\begin{matrix}I_{p}&0\\-D^{-1}C&I_{q}\end{matrix}}\right]\left[{\begin{matrix}(A-BD^{-1}C)^{-1}&0\\0&D^{-1}\end{matrix}}\right]\left[{\begin{matrix}I_{p}&-BD^{-1}\\0&I_{q}\end{matrix}}\right]\\[12pt]&=\left[{\begin{matrix}\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{matrix}}\right].\end{aligned}}}

C.f. matrix inversion lemma which illustrates relationships between the above and the equivalent derivation with the roles of A and D interchanged.

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:

${\displaystyle M^{-1}={\frac {1}{AD-BC}}\left[{\begin{matrix}D&-B\\-C&A\end{matrix}}\right]}$

provided that AD − BC is non-zero.

Moreover, the determinant of M is also clearly seen to be given by

${\displaystyle \det(M)=\det(D)\det(A-BD^{-1}C)}$

which generalizes the determinant formula for 2x2 matrices.

Application to solving linear equations

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

${\displaystyle Ax+By=a\,}$
${\displaystyle Cx+Dy=b\,}$

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 ${\displaystyle BD^{-1}}$ and then subtracting from the top equation one obtains

${\displaystyle (A-BD^{-1}C)x=a-BD^{-1}b.\,}$

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 ${\displaystyle Cx+Dy=b}$ one can solve for y. This reduces the problem of inverting a ${\displaystyle (p+q)\times (p+q)}$ 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

${\displaystyle \Sigma =\left[{\begin{matrix}A&B\\B^{T}&C\end{matrix}}\right],}$

where ${\displaystyle A\in \mathbb {R} ^{n\times n}}$ is the covariance matrix of X, ${\displaystyle C\in \mathbb {R} ^{m\times m}}$ is the covariance matrix of Y and ${\displaystyle B\in \mathbb {R} ^{n\times m}}$ is the covariance matrix between X and Y.

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

${\displaystyle \operatorname {Cov} (X\mid Y)=A-BC^{-1}B^{T}.}$
${\displaystyle \operatorname {E} (X\mid Y)=\operatorname {E} (X)+BC^{-1}(Y-\operatorname {E} (Y)).}$

If we take the matrix ${\displaystyle \Sigma }$ 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 ${\displaystyle \Sigma }$ also has a Wishart distribution.[citation needed]

Schur complement condition for positive definiteness and positive semidefiniteness

Let X be a symmetric matrix given by

${\displaystyle X=\left[{\begin{matrix}A&B\\B^{T}&C\end{matrix}}\right].}$

Let X/A be the Schur complement of A in X, i.e.

${\displaystyle X/A=C-B^{T}A^{-1}B,\,}$

and X/C be the Schur complement of C in X, i.e.

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

Then

• X is positive definite if and only if A and X/A are both positive definite:
${\displaystyle X\succ 0\Leftrightarrow A\succ 0,X/A=C-B^{T}A^{-1}B\succ 0}$.
• X is positive definite if and only if C and X/C are both positive definite:
${\displaystyle X\succ 0\Leftrightarrow C\succ 0,X/C=A-BC^{-1}B^{T}\succ 0}$.
• If A is positive definite, then X is positive semidefinite if and only if X/A is positive semidefinite:
${\displaystyle {\text{If}}}$ ${\displaystyle A\succ 0}$, ${\displaystyle {\text{then}}}$ ${\displaystyle X\succeq 0\Leftrightarrow X/A=C-B^{T}A^{-1}B\succeq 0}$.
• If C is positive definite, then X is positive semidefinite if and only if X/C is positive semidefinite:
${\displaystyle {\text{If}}}$ ${\displaystyle C\succ 0}$, ${\displaystyle {\text{then}}}$ ${\displaystyle X\succeq 0\Leftrightarrow X/C=A-BC^{-1}B^{T}\succeq 0}$.

The first and third statements can be derived[3] by considering the minimizer of the quantity

${\displaystyle u^{T}Au+2v^{T}B^{T}u+v^{T}Cv,\,}$

as a function of v (for fixed u).

Furthermore, since

${\displaystyle \left[{\begin{matrix}A&B\\B^{T}&C\end{matrix}}\right]\succ 0\Longleftrightarrow \left[{\begin{matrix}C&B^{T}\\B&A\end{matrix}}\right]\succ 0}$

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 semidefiniteness of X in terms of a generalized Schur complement.[1] Precisely,

• ${\displaystyle X\succeq 0\Leftrightarrow A\succeq 0,C-B^{T}A^{g}B\succeq 0,(I-AA^{g})B=0\,}$ and
• ${\displaystyle X\succeq 0\Leftrightarrow C\succeq 0,A-BC^{g}B^{T}\succeq 0,(I-CC^{g})B^{T}=0,}$

where ${\displaystyle A^{g}}$ denotes the generalized inverse of ${\displaystyle A}$.