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Quadratic form, law of inertia

by Gauss M.D.
Tags: form, inertia, quadratic
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Gauss M.D.
#1
Dec20-12, 06:49 AM
P: 150
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?
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pasmith
#2
Dec20-12, 07:05 AM
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Quote Quote by Gauss M.D. View Post
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?
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.
#3
Dec20-12, 07:14 AM
P: 150
Quote Quote by pasmith View Post
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]
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
#4
Dec20-12, 07:34 AM
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Quadratic form, law of inertia

Quote Quote by Gauss M.D. View Post
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?
Yes, but rearranging [itex]Q(x,y) = x(3x + 4y + 8z) + z(3z + 4y)[/itex] is not completing the square.
Frogeyedpeas
#5
Dec21-12, 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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