Properties of a special block matrix

ekkilop
Messages
29
Reaction score
0
Hi folks!

I've encountered the matrix below and I'm curious about its properties;

R=<br /> \begin{pmatrix}<br /> 0 &amp; N-S\\<br /> N+S &amp; 0<br /> \end{pmatrix}<br />

where R, N and S are real matrices, R is 2n by 2n, N is n by n symmetric and S is n by n skew-symmetric.

Clearly R is symmetric so the eigenvalues are real, but what else can be said about a matrix of this type? I checked through some literature but didn't really know what to look for. Surely the form is simple enough that it should have been studied.

In a special case, the elements of the rows of the matrix N+S sum to zero. Could this affect the properties somehow?

Any ideas would be much appreciated!
 
Physics news on Phys.org
Any matrix can be written as a symmetric plus a skew symmetric, so all you really have is a matrix of the form

R=<br /> \begin{pmatrix}<br /> 0 &amp; A\\<br /> A^t &amp; 0<br /> \end{pmatrix}<br />

The N and S aren't adding anything.

What are you doing with this matrix? Is there a specific type of problem you are trying to solve for example?
 
That's a fair point.
I was playing around with a different matrix - Hermitian and also symmetric about the anti-diagonal. Turns out that the eigenvectors are closely related to the eigenvectors of the matrix R above so I was curious about their structure. It seems reasonable that the upper and lower half of the eigenvectors should be closely related but the form largely depends on A I suppose.
But the form of R seems particularly neat so I thought perhaps it had some other interesting properties. Perhaps one could say what the determinant should be? Is it generally true that
det(R)=det(-A^{T}A)
 
ekkilop said:
Is it generally true that
det(R)=det(A^{T}A) [\tex]
<br /> That should be true for all block matrices (maybe with a different sign).<br /> As determinants are multiplicative, this can be simplified to det(R)=det(A)^2.
 
Thank you!
I think I shall have to return to the drawing board for a closer investigation :)
 

Similar threads

Back
Top