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1 vote

Determinant diagonal blocks compound matrix

There is a general theorem, which can be proved by using Schur's complement formula: let a matrix $A\in M_{pq}(k)$ be written blockwise, with blocks $A_{ij}\in M_q(k)$ for $1\le i,j\le p$. Assume that …
Denis Serre's user avatar
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10 votes

Determinant of block tridiagonal matrices

This is a fair example of the following theorem : let $A_{ij}\in M_r(k)$ be pairwise commuting matrices for $1\le i,j\le d$, and let $A\in M_{dr}(k)$ be the matrix whose $r\times r$ blocks are the $A_ …
Denis Serre's user avatar
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4 votes
Accepted

A closed-form expression for the inverse of a block-matrix

Say that $$B^{-1}=:\begin{pmatrix} b & X^T \\ Y & M \end{pmatrix}.$$ Then using Schur's complement formula (thanks to Nathaniel), $b=(x-{\bf1}^TA^{-1}{\bf1})^{-1}$ and $M=(A-x^{-1}{\bf11}^T)^{-1}$. Fr …
Denis Serre's user avatar
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11 votes
Accepted

Iterated calculation of determinants

The answer is Yes. This is done in Exercise 120 of my web page link text. You can replace $4$ and $2$ by numbers $n$ and $m$ with $m$ dividing $n$. Later. The required complement. Let me take the sit …
Denis Serre's user avatar
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9 votes
Accepted

One question on block-circulant matrices

The formula for the specific case is $$\det K=\det(A+B+C+D)\det(A-B+C-D)\det(A+iB-C-iD)\det(A-iB-C+iD).$$ More generally, for a block-circulant matrix with $n$ square blocks $A_0,\ldots,A_{n-1}$, the …
Denis Serre's user avatar
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