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Positive definite matrix bounded below

  1. Sep 7, 2011 #1
    1. The problem statement, all variables and given/known data

    Let [tex]A[/tex] be a positive definite [tex]n\times n[/tex] real matrix, [tex]b\in\mathbb{R}^n[/tex], and consider the quadratic polynomial [tex]Q(x)=\frac{1}{2}\langle{x, Ax\rangle}-\langle{b, x\rangle}[/tex]. Show that [tex]Q[/tex] is bounded below.

    2. The attempt at a solution

    I have to come up with a constant [tex]m[/tex] so that [tex]Q(x)\ge m[/tex] for all [tex]x\in\mathbb{R}^n.[/tex] I see that [tex]Q[/tex] looks a lot like a parabola. I know how to find the lower bound of a parabola opening upward but I don't know how to generalize this to quadratic forms.
     
    Last edited: Sep 7, 2011
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  3. Sep 7, 2011 #2

    micromass

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    What is a??
     
  4. Sep 7, 2011 #3
    Sorry, its supposed to be x.
     
  5. Sep 7, 2011 #4

    micromass

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    Let C be such that [itex]C^2=A[/itex] (why does C exist?) and let [itex]c=C^{-1}b[/itex] (why is C invertible?).

    Calculate

    [tex]\frac{1}{2}<Cx-c,Cx-c>[/tex]
     
  6. Sep 7, 2011 #5
    This is what I got but I don't know what it does

    [tex]\frac{1}{2}\langle{Cx-C^{-1}b, Cx-C^{-1}b\rangle} = \frac{1}{2}||Cx-C^{-1}b||\\
    =\frac{1}{2}||C^{-1}(Ax-b)||[/tex]

    [tex]C[/tex] exists because A is positive definite. [tex]C^{-1}[\tex] exists because A has no zero eigenvalues.
     
    Last edited: Sep 7, 2011
  7. Sep 7, 2011 #6

    micromass

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    Yes, so that shows that

    [tex]<Cx-C^{-1}b,Cx-X^{-1}b>\geq 0[/tex]

    Now try to evaluate this in another way... Use the linearity of the inproduct.
     
  8. Sep 10, 2011 #7
    This suggestion helped a lot. Thanks.
     
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