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Is the signature of this matrix zero?

  1. Jun 4, 2012 #1

    lavinia

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    Given are two square matrices of the same dimension, M and N.

    M is symmetric. N is non singular.

    From M and N form the symmetric matrix,

    M N
    N* 0

    Where N* is the transpose of N.

    Is the signature of this matrix necessarily zero? Counterexample?
     
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  3. Jun 5, 2012 #2

    AlephZero

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    My "engineer's intuition" says it is true, if you think about finding the extrema of x*Mx subject to the constraints Nx = 0, using Lagrange multipliers. The situations where this runs into trouble are when the constraints are not all independent, but that is excluded because N is non-singular.

    Basically, each eigenvalue of M is turned into a pair of eigenvalues of the augmented matrix, one positive and one negative.

    But my brain isn't working well enough to turn that idea into a proof right now - sorry!
     
  4. Jun 5, 2012 #3
    I think a topological argument works here.

    If you continuously deform such a matrix (within the constraints you mentioned), then the signature cannot change without an eigenvalue crossing 0. But that cannot happen without the determinant being zero. But such a matrix cannot have zero determinant because its rows are independent. In other words, the signature cannot change under a continuous deformation that maintains the given constraints on M and N.

    Every symmetric matrix can be continuously deformed to 0 (by scaling it down to 0). And GL(n) only has two connected components, so each non-singular matrix can be deformed to I or I' (which is I with one entry flipped to -1). So it suffices to check M=0, N=I, and M=0,N=I'.
     
  5. Jun 5, 2012 #4

    I like Serena

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    Pick for instance M=(0) and N=(1).

    Determinant=-1
    Trace=0
    Eigenvalues are +1 and -1.
     
  6. Jun 6, 2012 #5

    lavinia

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    Very cool argument. It took me a while to get it.

    I am going to try to generalize this to the matrix

    M T 0
    T* M N
    0 N 0

    where the conclusion now - I think - would be that the signature is this matrix is the signature of M. Can I just shrink T to zero?
     
    Last edited: Jun 6, 2012
  7. Jun 6, 2012 #6

    lavinia

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    This is the approach I have tried - more or less. I get a system of quadratic equations.
     
  8. Jun 6, 2012 #7

    lavinia

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    If T is shrunk towards zero the the matrix becomes singular only if M is singular so your proof goes through.
     
  9. Jun 6, 2012 #8
    Lets see... The determinant of this matrix will equal [itex] (-1)^n(\det M) (\det N)^2[/itex], where n is the block size. If M is non-singular, the determinant is never zero, so no signature changes could happen by shrinking T down to zero.

    If M is singular, the conclusion still sounds plausible, but I'm not sure whether the same proof could be adapted.
     
  10. Jun 9, 2012 #9

    lavinia

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    If M is diagonal then T can be shrunk to zero in the plane of non-zero eigen values.

    Choose a basis of eigen vectors for the copy of M that is in the upper left corner of the matrix.
     
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