Is Q a Positive Definite Matrix in this Mathematical Proof?

[EngWiPy]

Hi,

Suppose we have:

q_{ij}=\int_0^1x^{i+j}\,dx

can we prove that

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

is positive definite matrix? That is:

\mathbf{d}^T\mathbf{Q}\mathbf{d}>0

for all d?

Thanks in advance

 

[Landau]

q_{ij}=\int_0^1x^{i+j}\,dx

Here i,j are natural numbers between 1 and some n? If i+j+1 is never zero, then

q_{ij}=\frac{1}{i+j+1}

or am I missing something?

 

[Simon_Tyler]

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.

 

[AlephZero]

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.

 

[EngWiPy]

Here i,j are natural numbers between 1 and some n? If i+j+1 is never zero, then

q_{ij}=\frac{1}{i+j+1}

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

 

[AlephZero]

If you want to relate this to optimization and least squares fitting, then consider the problem of fitting a polynomial

P(x) = a_1 x + a_2 a^2 + \cdots + a_n x^n

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

\int_0^1 (P(x) - F(x))^2 dx

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?

 

[EngWiPy]

If you want to relate this to optimization and least squares fitting, then consider the problem of fitting a polynomial

P(x) = a_1 x + a_2 a^2 + \cdots + a_n x^n

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

\int_0^1 (P(x) - F(x))^2 dx

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 \mathbf{Q}. 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 \mathbf{a} is a strict relative minimum point?

Any other ideas?

Thanks

 

[AlephZero]

No, we are both trying to make this too complicated.

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

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

That's all there is to it.

 

[Simon_Tyler]

No, we are both trying to make this too complicated.

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

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!