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3 variable quadratic

  1. Feb 19, 2008 #1
    Hey guys.
    Im having a problem with deciphering positive definite and negative definite for this quadratic form i determined from a matrix.
    I dont quite understand how to evaluate if its pos def or neg def.
    From what I see in my notes, q(x,y) > 0 for all x not equal to 0. This doesnt help much as with some terms being squared and some not, im not sure if q will always be positive. I also dont know if there is anything different being a 3-variable quadratic. Anyways here it is:
    1. The problem statement, all variables and given/known data

    q(x, y, z) = 2x² - 2xy + y² - 8yz – 2xz + 5z²

    2. Relevant equations

    q(x,y) > 0 for all x not equal to 0

    Any help is appreciated
    Thanks

    Chris
     
  2. jcsd
  3. Feb 19, 2008 #2

    NateTG

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    Have you tried applying the quadratic formula?
     
  4. Feb 19, 2008 #3

    D H

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    What does positive definite mean in terms of the eigenvalues of the matrix?
     
  5. Feb 19, 2008 #4

    Dick

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    Before you do a lot of work, you could also just poke around. If v=(x,y,z)=(1,1,0) then q(v,v)=1. If w=(0,1,1), q(w,w)=(-2).
     
  6. Feb 19, 2008 #5
    hmm, well what im not sure of most is the statement "q(x,y) > 0 for all x not equal to 0"
    This was given for 2 variables, i have nothing for 3 variables. So in im not even sure if we are only taking into consideration what x is or if there are 2 variables to take consideration of in 3D. This is the only definition of positive definite i have :frown:
    I could do what Dick says by plug and play but im still not sure if the same definition holds for 3D....
     
  7. Feb 19, 2008 #6

    Dick

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    For two variables the statement is q(x,y)>0 for (x,y) not equal to (0,0). For three variables the statement is q(x,y,z)>0 for (x,y,z) not equal to (0,0,0). I wrote it the way I did because I usually think of a quadratic form as a function of two vectors, like the dot product.
     
  8. Feb 20, 2008 #7

    D H

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    To make Dick's suggestion blatantly obvious, setting x=1,y=1,z=0 yields q(x,y,z)=1 while setting x=0,y=1,z=1 yields q(x,y,z)=-2. What does that tell you regarding the question of the nature of the quadratic form?


    You can rewrite the expression q(x, y, z) = 2x² - 2xy + y² - 8yz – 2xz + 5z² as the matrix expression

    [tex]q(\mathbf x)
    = \mathbf x\cdot(\mathbf Q \mathbf x)
    = \mathbf x^T\mathbf Q \mathbf x[/tex]

    where [itex]\mathbf Q[/itex] is a symmetric matrix and [itex]\mathbf x[/itex] is the column vector

    [tex]\mathbf x = \bmatrix x\\y\\z\endbmatrix[/tex]

    The first form ([itex]\mathbf x\cdot(\mathbf Q \mathbf x)[/itex]) is apparently how Dick likes to view these forms. I prefer the second form ([itex]\mathbf x^T\mathbf Q \mathbf x[/itex]). It is just a matter of preference; the two expressions are equivalent.
     
  9. Feb 20, 2008 #8

    Dick

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    I prefer the second form as well. I think of the dot product as [itex]\mathbf x^T\mathbf I \mathbf x[/itex].
     
  10. Feb 20, 2008 #9
    Thank you guys. It appears to me that this quad is not pos or neg definite then. This is my 3rd option.
    Thanks guys

    Chris
     
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