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

Suppose p, q are nonnegative integers such that p + q > 0, and suppose A, B, C, D are respectively p × p, p × q, q × p, and q × q matrices of complex numbers. Let $${\displaystyle M=\left[{\begin{matrix}A&B\\C&D\end{matrix}}\right]}$$ so that M is a (p + q) × (p + q) matrix.

If D is invertible, then the Schur complement of the block D of the matrix M is the p × p matrix defined by $${\displaystyle M/D:=A-BD^{-1}C.}$$ If A is invertible, the Schur complement of the block A of the matrix M is the q × q matrix defined by $${\displaystyle M/A:=D-CA^{-1}B.}$$ 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^[1] who used it to prove Schur's lemma, although it had been used previously.^[2] Emilie Virginia Haynsworth was the first to call it the Schur complement.^[3] The Schur complement is a key tool in the fields of numerical analysis, statistics, and matrix analysis. The Schur complement is sometimes referred to as the Feshbach map after a physicist Herman Feshbach.^[4]

The Schur complement arises when performing a block Gaussian elimination on the matrix M. In order to eliminate the elements below the block diagonal, one multiplies the matrix M by a block lower triangular matrix on the right as follows: $${\displaystyle {\begin{aligned}&M={\begin{bmatrix}A&B\\C&D\end{bmatrix}}\quad \to \quad {\begin{bmatrix}A&B\\C&D\end{bmatrix}}{\begin{bmatrix}I_{p}&0\\-D^{-1}C&I_{q}\end{bmatrix}}={\begin{bmatrix}A-BD^{-1}C&B\\0&D\end{bmatrix}},\end{aligned}}}$$ where I_p denotes a p×p identity matrix. As a result, the Schur complement ${\displaystyle M/D=A-BD^{-1}C}$ appears in the upper-left p×p block.

Continuing the elimination process beyond this point (i.e., performing a block Gauss–Jordan elimination), $${\displaystyle {\begin{aligned}&{\begin{bmatrix}A-BD^{-1}C&B\\0&D\end{bmatrix}}\quad \to \quad {\begin{bmatrix}I_{p}&-BD^{-1}\\0&I_{q}\end{bmatrix}}{\begin{bmatrix}A-BD^{-1}C&B\\0&D\end{bmatrix}}={\begin{bmatrix}A-BD^{-1}C&0\\0&D\end{bmatrix}},\end{aligned}}}$$ leads to an LDU decomposition of M, which reads $${\displaystyle {\begin{aligned}M&={\begin{bmatrix}A&B\\C&D\end{bmatrix}}={\begin{bmatrix}I_{p}&BD^{-1}\\0&I_{q}\end{bmatrix}}{\begin{bmatrix}A-BD^{-1}C&0\\0&D\end{bmatrix}}{\begin{bmatrix}I_{p}&0\\D^{-1}C&I_{q}\end{bmatrix}}.\end{aligned}}}$$ Thus, the inverse of M may be expressed involving D⁻¹ and the inverse of Schur's complement, assuming it exists, as $${\displaystyle {\begin{aligned}M^{-1}={\begin{bmatrix}A&B\\C&D\end{bmatrix}}^{-1}={}&\left({\begin{bmatrix}I_{p}&BD^{-1}\\0&I_{q}\end{bmatrix}}{\begin{bmatrix}A-BD^{-1}C&0\\0&D\end{bmatrix}}{\begin{bmatrix}I_{p}&0\\D^{-1}C&I_{q}\end{bmatrix}}\right)^{-1}\\={}&{\begin{bmatrix}I_{p}&0\\-D^{-1}C&I_{q}\end{bmatrix}}{\begin{bmatrix}\left(A-BD^{-1}C\right)^{-1}&0\\0&D^{-1}\end{bmatrix}}{\begin{bmatrix}I_{p}&-BD^{-1}\\0&I_{q}\end{bmatrix}}\\[4pt]={}&{\begin{bmatrix}\left(A-BD^{-1}C\right)^{-1}&-\left(A-BD^{-1}C\right)^{-1}BD^{-1}\\-D^{-1}C\left(A-BD^{-1}C\right)^{-1}&D^{-1}+D^{-1}C\left(A-BD^{-1}C\right)^{-1}BD^{-1}\end{bmatrix}}\\[4pt]={}&{\begin{bmatrix}\left(M/D\right)^{-1}&-\left(M/D\right)^{-1}BD^{-1}\\-D^{-1}C\left(M/D\right)^{-1}&D^{-1}+D^{-1}C\left(M/D\right)^{-1}BD^{-1}\end{bmatrix}}.\end{aligned}}}$$ The above relationship comes from the elimination operations that involve D⁻¹ and M/D. An equivalent derivation can be done with the roles of A and D interchanged. By equating the expressions for M⁻¹ obtained in these two different ways, one can establish the matrix inversion lemma, which relates the two Schur complements of M: M/D and M/A (see "Derivation from LDU decomposition" in Woodbury matrix identity § Alternative proofs).

• 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.

• In general, if A is invertible, then

${\displaystyle {\begin{aligned}M&={\begin{bmatrix}A&B\\C&D\end{bmatrix}}={\begin{bmatrix}I_{p}&0\\CA^{-1}&I_{q}\end{bmatrix}}{\begin{bmatrix}A&0\\0&D-CA^{-1}B\end{bmatrix}}{\begin{bmatrix}I_{p}&A^{-1}B\\0&I_{q}\end{bmatrix}},\\[4pt]M^{-1}&={\begin{bmatrix}A^{-1}+A^{-1}B(M/A)^{-1}CA^{-1}&-A^{-1}B(M/A)^{-1}\\-(M/A)^{-1}CA^{-1}&(M/A)^{-1}\end{bmatrix}}\end{aligned}}}$

whenever this inverse exists.

• (Schur's formula) When A, respectively D, is invertible, the determinant of M is also clearly seen to be given by

${\displaystyle \det(M)=\det(A)\det \left(D-CA^{-1}B\right)}$, respectively

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

which generalizes the determinant formula for 2 × 2 matrices.

• (Guttman rank additivity formula) If D is invertible, then the rank of M is given by

${\displaystyle \operatorname {rank} (M)=\operatorname {rank} (D)+\operatorname {rank} \left(A-BD^{-1}C\right)}$

• (Haynsworth inertia additivity formula) If A is invertible, then the inertia of the block matrix M is equal to the inertia of A plus the inertia of M/A.
• (Quotient identity) ${\displaystyle A/B=((A/C)/(B/C))}$.^[5]
• The Schur complement of a Laplacian matrix is also a Laplacian matrix.^[6]

The Schur complement arises naturally in solving a system of linear equations such as^[7]

${\displaystyle {\begin{bmatrix}A&B\\C&D\end{bmatrix}}{\begin{bmatrix}x\\y\end{bmatrix}}={\begin{bmatrix}u\\v\end{bmatrix}}}$.

Assuming that the submatrix ${\displaystyle A}$ is invertible, we can eliminate ${\displaystyle x}$ from the equations, as follows.

${\displaystyle x=A^{-1}(u-By).}$

Substituting this expression into the second equation yields

${\displaystyle \left(D-CA^{-1}B\right)y=v-CA^{-1}u.}$

We refer to this as the reduced equation obtained by eliminating ${\displaystyle x}$ from the original equation. The matrix appearing in the reduced equation is called the Schur complement of the first block ${\displaystyle A}$ in ${\displaystyle M}$:

${\displaystyle S\ {\overset {\underset {\mathrm {def} }{}}{=}}\ D-CA^{-1}B}$.

Solving the reduced equation, we obtain

${\displaystyle y=S^{-1}\left(v-CA^{-1}u\right).}$

Substituting this into the first equation yields

${\displaystyle x=\left(A^{-1}+A^{-1}BS^{-1}CA^{-1}\right)u-A^{-1}BS^{-1}v.}$

We can express the above two equation as:

${\displaystyle {\begin{bmatrix}x\\y\end{bmatrix}}={\begin{bmatrix}A^{-1}+A^{-1}BS^{-1}CA^{-1}&-A^{-1}BS^{-1}\\-S^{-1}CA^{-1}&S^{-1}\end{bmatrix}}{\begin{bmatrix}u\\v\end{bmatrix}}.}$

Therefore, a formulation for the inverse of a block matrix is:

${\displaystyle {\begin{bmatrix}A&B\\C&D\end{bmatrix}}^{-1}={\begin{bmatrix}A^{-1}+A^{-1}BS^{-1}CA^{-1}&-A^{-1}BS^{-1}\\-S^{-1}CA^{-1}&S^{-1}\end{bmatrix}}={\begin{bmatrix}I_{p}&-A^{-1}B\\&I_{q}\end{bmatrix}}{\begin{bmatrix}A^{-1}&\\&S^{-1}\end{bmatrix}}{\begin{bmatrix}I_{p}&\\-CA^{-1}&I_{q}\end{bmatrix}}.}$

In particular, we see that the Schur complement is the inverse of the ${\displaystyle 2,2}$ block entry of the inverse of ${\displaystyle M}$.

In practice, one needs ${\displaystyle A}$ to be well-conditioned in order for this algorithm to be numerically accurate.

This method is useful in electrical engineering to reduce the dimension of a network's equations. It is especially useful when element(s) of the output vector are zero. For example, when ${\displaystyle u}$ or ${\displaystyle v}$ is zero, we can eliminate the associated rows of the coefficient matrix without any changes to the rest of the output vector. If ${\displaystyle v}$ is null then the above equation for ${\displaystyle x}$ reduces to ${\displaystyle x=\left(A^{-1}+A^{-1}BS^{-1}CA^{-1}\right)u}$, thus reducing the dimension of the coefficient matrix while leaving ${\displaystyle u}$ unmodified. This is used to advantage in electrical engineering where it is referred to as node elimination or Kron reduction.

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

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

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 ${\textstyle \Sigma }$:^[8]

${\displaystyle {\begin{aligned}\operatorname {Cov} (X\mid Y)&=A-BC^{-1}B^{\mathrm {T} }\\\operatorname {E} (X\mid Y)&=\operatorname {E} (X)+BC^{-1}(Y-\operatorname {E} (Y))\end{aligned}}}$

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]}

Let X be a symmetric matrix of real numbers given by $${\displaystyle X=\left[{\begin{matrix}A&B\\B^{\mathrm {T} }&C\end{matrix}}\right].}$$ Then

• If A is invertible, then X is positive definite if and only if A and its complement X/A are both positive definite:^[2]: 34

$${\displaystyle X\succ 0\Leftrightarrow A\succ 0,X/A=C-B^{\mathrm {T} }A^{-1}B\succ 0.}$$

• If C is invertible, then X is positive definite if and only if C and its complement X/C are both positive definite:

$${\displaystyle X\succ 0\Leftrightarrow C\succ 0,X/C=A-BC^{-1}B^{\mathrm {T} }\succ 0.}$$

• If A is positive definite, then X is positive semi-definite if and only if the complement X/A is positive semi-definite:^[2]: 34

$${\displaystyle {\text{If }}A\succ 0,{\text{ then }}X\succeq 0\Leftrightarrow X/A=C-B^{\mathrm {T} }A^{-1}B\succeq 0.}$$

• If C is positive definite, then X is positive semi-definite if and only if the complement X/C is positive semi-definite:

$${\displaystyle {\text{If }}C\succ 0,{\text{ then }}X\succeq 0\Leftrightarrow X/C=A-BC^{-1}B^{\mathrm {T} }\succeq 0.}$$

The first and third statements can be derived^[7] by considering the minimizer of the quantity $${\displaystyle u^{\mathrm {T} }Au+2v^{\mathrm {T} }B^{\mathrm {T} }u+v^{\mathrm {T} }Cv,\,}$$ as a function of v (for fixed u).

Furthermore, since $${\displaystyle \left[{\begin{matrix}A&B\\B^{\mathrm {T} }&C\end{matrix}}\right]\succ 0\Longleftrightarrow \left[{\begin{matrix}C&B^{\mathrm {T} }\\B&A\end{matrix}}\right]\succ 0}$$ 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.^[2] Precisely,

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

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

• Woodbury matrix identity
• Quasi-Newton method
• Haynsworth inertia additivity formula
• Gaussian process
• Total least squares