Quadratic form, law of inertia

Gauss M.D.

Getting ready for linear algebra exam. One question that I got right but not exactly sure why is this:

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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?
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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?

pasmith

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]

Gauss M.D.

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?

pasmith

Yes, but rearranging [itex]Q(x,y) = x(3x + 4y + 8z) + z(3z + 4y)[/itex] is not completing the square.

Frogeyedpeas

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