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Positive definite matrix bounded below

  • Thread starter Charles49
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Homework Statement



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.
 
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Answers and Replies

  • #2
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What is a??
 
  • #3
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Sorry, its supposed to be x.
 
  • #4
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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]
 
  • #5
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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.
 
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  • #6
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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.
 
  • #7
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This suggestion helped a lot. Thanks.
 

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