# Schur's Complement and Inverse of Block Matrices

Assume that we are given a block matrix of the form: $$M = \left[ \begin{array}{cc} A & b \\ b^T & c \\ \end{array} \right]$$ where $b$ is a column vector. and $c$ is a scalar.

Schur's complement of $A$ in $M$ is given by: $$s = c - b^T A^{-1} b$$ Assume that we know $A$ is invertible and we want to check if $M$ is invertible. Is it true that $M$ is invertible if, and only if $s$ is different than zero?

Thank you

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Yes.

Under the assumption that $A^{-1}$ exists, then

$\det (M) = \det (A) s.$

Since $\det(A) \neq 0$, $\det (M) \neq 0$ if and only if $s \neq 0$. Thus $M$ is inveritible if and only if $s \neq 0$.

This theorem on block determinants is standard material that can be found in many linear algebra textbooks. See for example Carl Meyer's textbook, "Matrix Analysis and Linear Algebra."

The proof is pretty simple- write $M$ as

$M=\left[ \begin{array}{cc} I & 0 \\ b^{T}A^{-1} & 1 \\ \end{array} \right] \left[ \begin{array}{cc} A & b \\ 0 & c-b^{T}A^{-1}b \\ \end{array} \right]$

Then apply the product rule and the rule for determinants of (block) diagonal matrices.

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Mind you, it's not a very good idea to use this in computational practice to check for invertibility of $M$. – Brian Borchers Dec 20 '12 at 18:17
I think practicality hinges on the difficulty of solving $Au=b$, from which computation of $s$ will easily follow. The phrase bordered matrix may be a useful search term. – hardmath Dec 23 '12 at 15:39
Thank you for you reply. You mention that it is not a very good idea to use this in computational practice to check for invertibility of $M$. It would obviously be hard to get $s$ = exact zero, but at least $s$ should in the order of $10^{10}$ or even less if $M$ is singular, right? Can't this be used as a check? I can't seem to find a 100% reliable singularity check. – Mohammad Fawaz Dec 23 '12 at 22:22
The determinant is a very bad measure of singularity in practice, because it isn't scale invariant. For example, suppose that I tell you a matrix $M$ has $\det(M)=1.0 \times 10^{-15}$. You might argue that this means the matrix is singular or nearly singular. Now, let me tell you that this particular matrix is of size 15 by 15, and $M=0.1I$. Clearly, $M$ is nowhere near singular. The proper measure of singularity to use is the condition number, but this requires some knowledge of the eigenvalues of $M$. – Brian Borchers Dec 25 '12 at 3:31