Schur complement

Not to be confused with the Schur complement method in numerical analysis..

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

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

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

$A-BD^{-1}C.\,$

It 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]

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

$L=\left[\begin{matrix} I_p & 0 \\ -D^{-1}C & I_q \end{matrix}\right].$

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

\begin{align} 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{align}

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

\begin{align} \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{align}

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

\begin{align} & {} \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-B D^{-1} C \right)^{-1} & -\left(A-B D^{-1} C \right)^{-1} B D^{-1} \\ -D^{-1}C\left(A-B D^{-1} C \right)^{-1} & D^{-1}+ D^{-1} C \left(A-B D^{-1} C \right)^{-1} B D^{-1} \end{matrix} \right]. \end{align}

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:

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

provided that AD − BC is non-zero.

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

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

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

$Ax + By = a \,$
$Cx + Dy = b \,$

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

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

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

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

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

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

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

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

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

Let X be a symmetric matrix given by

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

Let S be the Schur complement of A in X, that is:

$S= C - B^T A^{-1} B . \,$

Then

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

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

$u^T A u + 2 v^T B^T u + v^T C v, \,$

as a function of v (for fixed u).