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Write a linear system as $Ax+By=f$ and $Cx+Dy=g$, with $A$ an invertible matrix. Solving the first equation for $x$ and substituting in the second givesThus the Schur complement is the coefficient matrix remaining after the $x$ variables have been eliminated. When it is an invertible matrix, solve the reduced linear system for $y$ and recover $x$. In computation, expressions such as $A^{-1}B$ mean solving with each column of $B$, using the same LU decomposition or Cholesky decomposition when applicable.

For the example under block matrix inverse and right-hand side $(3,1,4{)}^T$, take $f=(3,1{)}^T$ and $g=4$. Since $C{A}^{-1}f=13/6$, the reduced equation is $(19/6)y=11/6$. Back substitution givesThe original equations return $3$, $1$ and $4$, respectively. The reduction retained the influence of $x_1,x_2$ on $y$ through the correction $C{A}^{-1}B$.

A short proof outline also explains whether $M$ is an invertible matrix: the exact matrix factorizationhas invertible matrices on the outside. Therefore $M$ is an invertible matrix precisely when the central block diagonal matrix is one. Reversing the factors gives the block matrix inverse, while taking determinants gives the Schur complement determinant identity.

A block matrix inverse expresses the matrix inverse in terms of smaller matrices. Partition a square matrix over a field asIf $A$ is an invertible matrix, define its Schur complement $S=D-C{A}^{-1}B$. Then $M$ is an invertible matrix exactly when $S$ is an invertible matrix, andThe order of the matrix products matters. In particular, the lower-right block is $S^{-1}$, rather than generally $D^{-1}$: the coupling through $B$ and $C$ changes the response.^[ref]

For an original example, takeThe Schur complement is $S=5-1/2-4/3=19/6$. Substitution gives the concise answerMultiplying in either order gives the identity matrix; the lower-right entry is $6/19$, whereas $D^{-1}=1/5$. The difference records the feedback through the two eliminated variables.

Often the desired output is a solution of a linear system, rather than every entry of a matrix inverse. Block elimination solves with $A$ and then with $S$, recovering the eliminated variables by back substitution. Reusing an LU decomposition of $A$ avoids explicitly forming $A^{-1}$. A favourable block diagonal matrix or tridiagonal matrix block can make this useful; merely partitioning an unstructured dense matrix does not guarantee a faster solve.^[ref]

An invertible matrix need not have an invertible chosen diagonal block. For example,has singular matrices as both one-by-one diagonal blocks. Neither diagonal-block formula applies in that partition. Interchanging the two rows produces the identity matrix and a usable pivot element. This is why a failed block pivot is not proof that the full matrix is a singular matrix; Pivoted LU decomposition changes the order before eliminating.

Conversely, invertible diagonal blocks do not suffice: $M=\left(\begin{array}{cc}1& 1\\ 1& 1\end{array}\right)$ has $A=D=1$ but Schur complement $S=0$, so it is a singular matrix. Both hypotheses in the block matrix inverse formula have work to do.

If a linear smoother has fitted vector $\hat{Y}=HY$, its residual after fitting without observation $i$ isThe identity follows from a block matrix inverse or a rank-one inverse update.

The leave-one-out residual identity for a linear smoother, obtained from the block matrix inverse or the Sherman–Morrison formula, isHence

Compute once the spectral decomposition $K=U\mathrm{diag}(d_1,\dots ,d_n)U^T$ in $O(n^3)$ operations and the vector $U^{T}Y$ in $O(n^2)$. For each ${\lambda }_{\ell }$, setThen computeBoth calculations take $O(n^2)$ operations per tuning parameter, after which the displayed leave-one-out formula costs $O(n)$. All $L$ scores therefore require $O(n^3+L{n}^2)$ operations.

For a block matrix $M=\left(\begin{array}{cc}A& B\\ C& D\end{array}\right)$ with $A$ an invertible matrix, the Schur complement of $A$ in $M$ is $S=D-C{A}^{-1}B$. It is the effective coefficient matrix for the remaining variables after block elimination. The definition requires invertibility of $A$, but not of $S$ or $M$. Its size is that of $D$, even when $B$ and $C$ are rectangular matrices.

In the original example under block matrix inverse, $D=5$ becomes $S=19/6$ because changing the remaining variable also changes the two variables solved through $A$. This is what makes a Schur complement more informative than simply deleting rows and columns. It retains their mediated effect, and its matrix inverse is the corresponding block of $M^{-1}$ when the block matrix inverse exists.

The same elimination appears when minimizing a quadratic form. The Schur complement as a minimized quadratic form gives a geometric example; the Schur complement determinant identity gives an exact test for an invertible matrix.

When $A$ is an invertible matrix, the factorization in block elimination has outer triangular factors with determinant one. Hence $\det M=\det A\det S$, without a symmetry assumption. For the original example under block matrix inverse, this gives $\det M=(2\cdot 3)(19/6)=19$. An invertible block $A$ therefore reduces the question of whether the full matrix is an invertible matrix to the smaller Schur complement; a singular $S$ makes $M$ a singular matrix.