Determinant of a block matrix of a specific form

**JustMeDK** · October 25th 2012, 03:39 AM

Let $m_{1},m_{2}$ be two $n\times n$ matrices, and consider the $2n\times 2n$ block-matrix $M=\begin{pmatrix}m_{1} & -m_{2} \\ m_{2} & m_{1}\end{pmatrix}.$ Does there exist a formula for the determinant of $M$ in the general case where no assumptions on $m_{1}$ and $m_{2}$ are being made? The formulas for the determinants of block matrices given in Determinant - Wikipedia, the free encyclopedia or http://www.mth.kcl.ac.uk/~jrs/gazette/blocks.pdf, Eqs. (13)-(16) for the latter, are of no use, as they make assumptions of either invertibility or commutativity.

**chiro** · October 25th 2012, 03:46 AM

Hey JustMeDK.

Are you familiar with multi-linear algebra and tensors products? The reason I ask is that if there is a relationship between the determinants of the tensor products with respect to the determinant of the individual spaces then you can use this result.

It should look something like a set cardinality argument where |AXB| = |A||B|

**JustMeDK** · October 25th 2012, 03:50 AM

Hi chiro

Yes, having a masters degree in physics, I'm familiar with multilinear algebra and tensor products. So please elaborate a little more.

**chiro** · October 25th 2012, 03:59 AM

This might help (also scroll down to determinant properties):

Kronecker product - Wikipedia, the free encyclopedia

**JustMeDK** · October 25th 2012, 04:11 AM

Thanks for your link, although I was previously acquainted with that wiki-page. Unless I'm fundamentally mistaken, the determinantal formula of a tensor product as there given seems to be of no use, though, for the matrix $M$ as given above is a sum of two Kronecker products:

$M=\begin{pmatrix}1 & 0 \\ 0 & 1\end{pmatrix}\otimes m_{1}+\begin{pmatrix}0 & -1 \\ 1 & 0\end{pmatrix}\otimes m_{2}.$

**chiro** · October 25th 2012, 04:42 AM

The other thing relates to use row reduction operations to get the determinants but this will require you to do a bit of grunt-work.

For example det(AB) = det(A)det(B) for square matrices of same size and doing a row reduction and balancing it allows you to get multiplication of XM where X is the row reduction matrix to balance the reduction and M is the result of row-reducing.

You can then use these properties to show what they will be.

I haven't done the gruntwork myself, but I think this would be a starting point if you can't find an existing result.

**JustMeDK** · October 25th 2012, 05:03 AM

Please forgive me, but I do not think that row-reduction is the way forward. What I would like is to know whether or not there exists a formula for $det(M)$ in terms of $m_{1}$ and $m_{2}$ as 'black boxes', using determinants and/or traces, or whatever, of these, without assorting explicitly to their (completely unconstrained) entries. For doing the latter would, I believe, become completely untenable for higher dimensionalities; what I need is a formula valid for any value of $n$ .

**chiro** · October 25th 2012, 07:59 PM

Is it possible to consider an eigen-decomposition for your case given that the determinant is the product of the eigen-values?

**JustMeDK** · October 25th 2012, 10:24 PM

I don't think that is pratically doable because, for general $n$ , it corresponds to solving a characteristic polynomial of degree $2n$ in $2n^{2}$ free parameters.

**chiro** · October 25th 2012, 10:38 PM

Are there ways to do calculation given in block form (i.e. given blocks and eigenvalues, is there a generalization given a particular partitioned system)?

I am aware that partitioned systems are studied but personally don't know any results that have been obtained.

**JustMeDK** · October 25th 2012, 11:01 PM

Minutes ago I discovered the article http://arxiv.org/pdf/1112.4379v1.pdf on the first page of which the following interesting statement occurs: 'If neither inverse exists, the notion of generalized inverses must be employed', followed by three references. I will try to look up these references later today when visiting the university library.

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