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Quadratic form, law of inertia

  1. Dec 20, 2012 #1
    Getting ready for linear algebra exam. One question that I got right but not exactly sure why is this:

    Consider the quadratic form

    Q(x,y,z) = 3x^2 + 3z^2 + 4xy + 4xy + 8xz

    a) Decide if Q is positive definite, indefinite, etc.

    b) What point on the surface Q = 1 lies closest to the origin and what is that distance?

    I computed the eigenvalues and got -1, -1 and 8, i.e. indefinite. But when just completing the square, there is only two positive terms: x(3x + 4y + 8z) + z(3z + 4y). How does this mesh with Sylvesters law of inertia?

    Also, this form has got to be some kind of hyperboloid or something. So how can I know if the point associated with 1/sqrt(8) is actually on the surface? Since we're dealing with hyperbolas and not ellipses, that isn't always the case, is it?
  2. jcsd
  3. Dec 20, 2012 #2


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    That's not the result of completing the square. If you complete the squares in x, y and z in that order you should get
    [tex]3\left(x + \frac23y + \frac43z\right)^2 - \frac43\left(y + \frac12z\right)^2 - 2z^2[/tex]
  4. Dec 20, 2012 #3
    But I thought the law indicated that no matter how you complete the square, the number of positive and negative terms will always be the same?
  5. Dec 20, 2012 #4


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    Yes, but rearranging [itex]Q(x,y) = x(3x + 4y + 8z) + z(3z + 4y)[/itex] is not completing the square.
  6. Dec 21, 2012 #5
    out of curiosity what level of linear algebra is this? Because I just finished my course and we never covered this haha. Though we did talk about eigenvectors
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