- #1

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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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- Thread starter EngWiPy
- Start date

- #1

- 1,367

- 61

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

- #2

Landau

Science Advisor

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

[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.

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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- #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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[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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[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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[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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[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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