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Proving the inertia of a quadratic surface?

  1. Mar 22, 2009 #1
    1. The problem statement, all variables and given/known data

    Let F(x, y, z) be a quadratic form and A its associated matrix. The inertia of A, denoted in(A), is defined as the triple in(A) = (n1, n2, n3) where n1, n2, n3 denotes the number of positive, negative, and zero eigenvalues of A respectively. Prove the following:

    If in(A) = (3, 0, 0) then the quadratic is an ellipsoid.
    If in(A) = (2, 0, 1) then the quadratic is an elliptic paraboloid.
    If in(A) = (2, 1, 0) then the quadratic is a hyperboloid of one sheet.
    If in(A) = (1, 2, 0) then the quadratic is a hyperboloid of two sheets.
    If in(A) = (1, 1, 1) then the quadratic is a hyperbolic paraboloid.
    If in(A) = (1, 0, 2) then the quadratic is a parabolic cylinder.

    2. Relevant equations



    3. The attempt at a solution

    I thought that the number of eigenvalues has some direct connection with the defined equations for all of those surfaces. For example, the a, b, and c constants are all positive in the equation of an ellipsoid, so its inertia is (3, 0, 0). But, then what about, for example, the elliptic paraboloid? It's equation is z = x^2/a^2 + y^2/b^2 so it should have 2 positive eigenvalues. Where does the one negative come from? I am having trouble picturing the connection between eigenvalues and these surfaces.
     
  2. jcsd
  3. Mar 22, 2009 #2

    tiny-tim

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    Hi cookiesyum! :smile:
    No, it's a zero eigenvalue (along the z-axis). :wink:
     
  4. Mar 22, 2009 #3
    Oh, I see! Thanks. So the pattern makes sense, but how would I go about proving it?
     
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