Positive Definite Matrix

  • Thread starter EngWiPy
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  • #1
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Main Question or Discussion Point

Hi,

Suppose we have:

[tex]q_{ij}=\int_0^1x^{i+j}\,dx[/tex]

can we prove that

[tex]\mathbf{Q}=[q_{ij}][/tex]

is positive definite matrix? That is:

[tex]\mathbf{d}^T\mathbf{Q}\mathbf{d}>0[/tex]

for all d?

Thanks in advance
 

Answers and Replies

  • #2
Landau
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[tex]q_{ij}=\int_0^1x^{i+j}\,dx[/tex]
Here i,j are natural numbers between 1 and some n? If i+j+1 is never zero, then

[tex]q_{ij}=\frac{1}{i+j+1}[/tex]

or am I missing something?
 
  • #3
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The result given by Landau is called the http://en.wikipedia.org/wiki/Hilbert_matrix" [Broken]. It's a famous example of an ill-conditioned matrix. The wiki page linked to lists its properties.

As for a proof that it's positive definite. I think maybe the easiest (almost trivial) way would be to use "[URL [Broken] criterion[/URL] and induction.
 
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  • #4
AlephZero
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Another easy proof is to use the fact that if every 2x2 submatrix of a matrix M is positive definite, then M is positive definite.
 
  • #5
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Here i,j are natural numbers between 1 and some n? If i+j+1 is never zero, then

[tex]q_{ij}=\frac{1}{i+j+1}[/tex]

or am I missing something?
You are absolutely right. Can you go further?

Thank you Simon_Tyler and AlephZero for your replies, but I think these methods are advanced somewhat. I am taking this first course in optimization, and we use the method I mentioned in the first post.

Regards
 
  • #6
AlephZero
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If you want to relate this to optimization and least squares fitting, then consider the problem of fitting a polynomial

[tex]P(x) = a_1 x + a_2 a^2 + \cdots + a_n x^n[/tex]

to an arbitrary function [itex]F(x)[/itex] over the interval [itex][0,1][/itex]. Minimize

[tex]\int_0^1 (P(x) - F(x))^2 dx[/tex]

and your Hilbert matrix will appear. I can't remember much general optimization theory, but can you use this to prove the Hillbert matrix is positive definite?
 
  • #7
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If you want to relate this to optimization and least squares fitting, then consider the problem of fitting a polynomial

[tex]P(x) = a_1 x + a_2 a^2 + \cdots + a_n x^n[/tex]

to an arbitrary function [itex]F(x)[/itex] over the interval [itex][0,1][/itex]. Minimize

[tex]\int_0^1 (P(x) - F(x))^2 dx[/tex]

and your Hilbert matrix will appear. I can't remember much general optimization theory, but can you use this to prove the Hillbert matrix is positive definite?
Yes I know, and from this problem exactly I got the matrix [tex]\mathbf{Q}[/tex]. I did the first order necessary conditions, and moved to the second order conditions and stuck at the point at hand, which is: is Q a positive definite matrix? which means, is our solution of [tex]\mathbf{a}[/tex] is a strict relative minimum point?

Any other ideas?

Thanks
 
  • #8
AlephZero
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No, we are both trying to make this too complicated.

[tex]\int_0^1 [P(x)]^2 dx > 0[/tex]

for all possible values of the a's, except when all the a's are zero.

That's all there is to it.
 
  • #9
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No, we are both trying to make this too complicated.

[tex]\int_0^1 [P(x)]^2 dx > 0[/tex]

for all possible values of the a's, except when all the a's are zero.

That's all there is to it.
It's pretty obvious when you put it like that!
 

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