Properties of a special block matrix

  • Thread starter ekkilop
  • Start date
  • #1
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Hi folks!

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

[tex] R=
\begin{pmatrix}
0 & N-S\\
N+S & 0
\end{pmatrix}
[/tex]

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!
 

Answers and Replies

  • #2
Office_Shredder
Staff Emeritus
Science Advisor
Gold Member
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Any matrix can be written as a symmetric plus a skew symmetric, so all you really have is a matrix of the form

[tex] R=
\begin{pmatrix}
0 & A\\
A^t & 0
\end{pmatrix}
[/tex]

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?
 
  • #3
29
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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
[itex] det(R)=det(-A^{T}A) [/itex]
 
  • #4
35,628
12,165
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.
 
  • #5
29
0
Thank you!
I think I shall have to return to the drawing board for a closer investigation :)
 

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