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Quadratic form, law of inertia 
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#1
Dec2012, 06:49 AM

P: 149

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
Dec2012, 07:05 AM

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P: 945

[tex]3\left(x + \frac23y + \frac43z\right)^2  \frac43\left(y + \frac12z\right)^2  2z^2[/tex] 


#3
Dec2012, 07:14 AM

P: 149




#4
Dec2012, 07:34 AM

HW Helper
Thanks
P: 945

Quadratic form, law of inertia



#5
Dec2112, 03:38 PM

P: 80

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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