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Properties of a special block matrix

  1. May 5, 2013 #1
    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!
     
  2. jcsd
  3. May 5, 2013 #2

    Office_Shredder

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    Staff Emeritus
    Science Advisor
    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

    [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?
     
  4. May 6, 2013 #3
    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]
     
  5. May 6, 2013 #4

    mfb

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    2016 Award

    Staff: Mentor

    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.
     
  6. May 6, 2013 #5
    Thank you!
    I think I shall have to return to the drawing board for a closer investigation :)
     
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