Proving the inertia of a quadratic surface?

cookiesyum
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Homework Statement



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.

Homework Equations





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.
 
on Phys.org
Hi cookiesyum! :smile:
cookiesyum said:
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?

No, it's a zero eigenvalue (along the z-axis). :wink:
 
tiny-tim said:
Hi cookiesyum! :smile:No, it's a zero eigenvalue (along the z-axis). :wink:

Oh, I see! Thanks. So the pattern makes sense, but how would I go about proving it?
 

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