This is a tiny followup of a previous post on nonlinear formulas in linear algebra. Let us consider a block matrix ${M}$ of size ${(n+m)\times(n+m)}$ of the form

$M= \begin{pmatrix} A & B \\C & D \end{pmatrix}$

where ${A,B,C,D}$ are ${n\times n}$, ${n\times m}$, ${m\times n}$, ${m\times m}$. If ${D}$ is invertible then

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

This follows immediately from the identity (mentioned in Wikipedia)

$\begin{pmatrix} A & B \\ C & D \end{pmatrix} \begin{pmatrix} I & 0 \\ -D^{-1}C & I \end{pmatrix} = \begin{pmatrix} A-BD^{-1}C & B \\ 0 & D \end{pmatrix}$

(the determinant of a block triangular matrix is the product of the determinants of its diagonal blocks). If ${m=n}$ and if ${C},{D}$ commute then ${\det(M)=\det(AD-BC)}$. Note that ${A-BD^{-1}C}$ is the Schur complement of ${A}$ in ${M}$. Similar formulas are derived in arXiv:1112.4379 for the determinant of ${nN\times nN}$ block matrices formed by ${N^2}$ blocks of size ${n\times n}$.

1. Gérard Letac 2012-10-22

The paper by Philip Powell recommended at the end is naive and is nothing but an ordinary Cholevsky decomposition of a matrix by blocks…

2. Minghua 2013-01-09

Any similar formula for permanent function instead of determinant?

3. Nickie 2018-08-08

I have applied the analysis in Powell’s paper to a block companion matrix and the result I get is inconsistent with numerical results obtained in MATLAB, is Powell’s method correct?!

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