Is Q a Positive Definite Matrix in this Mathematical Proof?

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Discussion Overview

The discussion revolves around the question of whether the matrix \(\mathbf{Q} = [q_{ij}]\), defined by the integral \(q_{ij} = \int_0^1 x^{i+j} \, dx\), is a positive definite matrix. Participants explore various approaches to prove this property, considering its implications in optimization and least squares fitting.

Discussion Character

  • Exploratory
  • Technical explanation
  • Mathematical reasoning
  • Debate/contested

Main Points Raised

  • Some participants propose that if \(i+j+1\) is never zero, then \(q_{ij} = \frac{1}{i+j+1}\) holds true.
  • One participant mentions the Hilbert matrix as a well-known example of an ill-conditioned matrix and suggests using a specific criterion and induction for proving positive definiteness.
  • Another participant suggests that if every 2x2 submatrix of a matrix is positive definite, then the matrix itself is positive definite.
  • There is a discussion about relating the matrix to optimization problems, specifically in the context of fitting a polynomial to a function over the interval [0,1].
  • One participant states that the integral \(\int_0^1 [P(x)]^2 dx > 0\) for all non-zero coefficients \(a\) implies a straightforward argument for positive definiteness.
  • Some participants express uncertainty about the complexity of the problem and suggest that the reasoning might be more straightforward than initially thought.

Areas of Agreement / Disagreement

Participants express differing views on the complexity of proving the positive definiteness of the matrix \(\mathbf{Q}\). While some suggest advanced methods, others believe a simpler approach suffices. No consensus is reached on the best method to prove the property.

Contextual Notes

Participants note that the discussion depends on the definitions of the variables involved and the assumptions about the indices \(i\) and \(j\). There are also unresolved mathematical steps in the proposed proofs.

EngWiPy
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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
 
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S_David said:
[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?
 
The result given by Landau is called the http://en.wikipedia.org/wiki/Hilbert_matrix" . 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 criterion[/URL] and induction.
 
Last edited by a moderator:
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.
 
Landau said:
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
 
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?
 
AlephZero said:
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
 
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.
 
AlephZero said:
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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