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Hermitian matrix with negative eigenvalue

  1. Mar 18, 2010 #1
    1. The problem statement, all variables and given/known data
    Hello,
    I have the following problem:
    Suppose A is a hermitian matrix and it has eigenvalue [tex]\lambda[/tex] <=0. Show that A is not positive definite i.e there exists vector v such that (v^T)(A)(v bar) <=0

    3. The attempt at a solution
    Let w be an eigenvetor we have the following equalities which are equivalent:
    Aw=[tex]\lambda[/tex]w
    A(w bar) = [tex]\lambda[/tex] (w bar) (i am not sure about this equality)
    (w^t)(A)(w bar) = (w^t)[tex]\lambda[/tex] (w bar)
    [(w^t)(A)(w bar)]^T = [(w^t)[tex]\lambda[/tex] (w bar)]^T
    (w bar)^T*(A)^T*w = (w bar)^T*[tex]\lambda[/tex]*w
    [(w bar)^T*(A)^T*w]bar = [(w bar)^T*[tex]\lambda[/tex]*w]bar
    (w^T)(A bar)^T(w bar) = (w^T)[tex]\lambda[/tex](w bar)
    (w^T)A(w bar) = (w^T)[tex]\lambda[/tex](w bar)

    but i cannot prove why (w^T)[tex]\lambda[/tex](w bar) is negative, assuming taht the 2ndd equality is true.

    Thank you
     
  2. jcsd
  3. Mar 18, 2010 #2

    Dick

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    Well, (w^T)(w bar) is positive, isn't it? It's sum of w_i*(w_i bar), right?
     
  4. Mar 19, 2010 #3
    I'm not to sure what you mean by w_i?

    Also, I'm still not completely convinced that
    Aw=[tex]\lambda[/tex]w is equivalent to A(w bar) = [tex]\lambda[/tex](w bar).
    Is it generally true that the conjugate is also an eigenvector?

    Thanks
     
  5. Mar 19, 2010 #4

    Dick

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    w_i is the i component of the vector w. No, you can't go from Aw=lambda*w to A(w bar)=lambda (w bar). What you can do is take conjugate transpose of both sides. (w bar)^T (A bar)^T=((lambda bar) (w bar)^T). Do you see how to get what you want from there?
     
  6. Mar 19, 2010 #5
    yea if we start off with (w bar)^T(A bar)^T then this is equivalent to the following(I am only looking at the left side of the equation)
    (w bar)^T(A bar)^T(w bar)
    [(w bar)^T(A bar)^T(w bar)]^T since it is a 1x1 matrix
    (w bar)^T(A bar)(w)
    [(w bar)^T(A bar)(w)]bar
    (w^T)(A)(w bar)

    and on the left side we're left with [tex]\lambda[/tex](w^T)(w bar) which is negative, hence the left side is negative and therefore A is not positive definite
    Correct?
     
  7. Mar 19, 2010 #6

    Dick

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    No, not correct. I really don't know how you are justifying some of those steps. Can you show what happens to the right side as well? And you have to use that A is hermitian, otherwise the statement isn't true. What does it mean for A to be Hermitian?
     
  8. Mar 19, 2010 #7
    So, far we have Aw=[tex]\lambda[/tex]w
    so by taking the transpose and getting the conjugate we get
    (w bar)^T(A bar)^T = (w bar)^T ([tex]\lambda[/tex] bar)^T but A is hermitian and lamba is a real number therefore we get

    (w bar)^T(A) = [tex]\lambda[/tex] (w bar)^T

    now if I want to arrive at w^TA(w bar) on the right side i will have to either add a w factor in which case i can:
    -get the transpose of the whole equation, at which point i will get an extra ^T on the A. (w^T)(A)^T(w bar)
    -or i can but a bar over the whole thing where i would be left with an extra bar on the A.(w^T)(A bar)(w bar)
    I cannot quite fix the A in order to get the correct equation. Is there some kind of property that I am missing here?

    Thanks again
     
  9. Mar 19, 2010 #8

    Dick

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    You just need 'some vector' v. v doesn't have to be the same as w. How about picking v=(w bar)?
     
  10. Mar 19, 2010 #9
    Ah yes I didn't think of that
    so we have (w bar)^T A = (lambda)(w bar)^T
    replacing w bar by v and then multiplying both sides by (v bar) we get
    (v^T)(A)(v bar) = (lambda)(v^T)(v bar) but (v^T)(v bar)= ((z1(z1 bar) + ... + zn(zn bar)) which is positive hence (lambda)(v^T)(v bar) is negative which is what we wanted to show.

    right?
     
  11. Mar 19, 2010 #10

    Dick

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    Right.
     
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