# Properties of a special block matrix

Hi folks!

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

$$R= \begin{pmatrix} 0 & N-S\\ N+S & 0 \end{pmatrix}$$

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!

Office_Shredder
Staff Emeritus
Gold Member
Any matrix can be written as a symmetric plus a skew symmetric, so all you really have is a matrix of the form

$$R= \begin{pmatrix} 0 & A\\ A^t & 0 \end{pmatrix}$$

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
$det(R)=det(-A^{T}A)$

mfb
Mentor
Is it generally true that
[tex] det(R)=det(A^{T}A) [\tex]
That should be true for all block matrices (maybe with a different sign).
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 :)