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        general >> wanted >> operators
        
(Message started by: MonicaMath on Nov 26^th, 2009, 11:11am)

Title: operators
Post by MonicaMath on Nov 26^th, 2009, 11:11am

Hi all,
can anyone help me !!

I wanna find the adjoint of the linear first order differential operator D
on polynomials of degree at most m from a vector space P[x]:
D(p(x)) = p'(x)

any help ?? any resources ???

with the inner product is defined as:

<p1(x), p2(x)> = SUM_k=0 ^m {a_k d_k*} , where dk* means conjugate of dk

p1(x)= SUM k=0^m (a_k x^k), p2(x)= SUM k=0^m (d_k x^k)

Title: Re: operators
Post by Obob on Nov 26^th, 2009, 12:02pm

The space of polynomials of degree at most m has as a basis the polynomials 1, x, x^2, x^3,...,x^m, and this is an orthonormal basis for the inner product you've written down.

Write down the operator D as a matrix in terms of this basis. Then the adjoint is given by the conjugate transpose of that matrix.

Title: Re: operators
Post by MonicaMath on Nov 26^th, 2009, 11:28pm

Ok,

The matrix representation for the operator D is:

0 1 0 0 0 ....... 0
0 0 2 0 0 ........ 0
0 0 0 3 0 ........ 0
.
.
.
0 0 0 0 0 ...... m
0 0 0 0 0 ...... 0

right ??

so now I take transpose !!

and if so, how I can form the adjoint D* from the resulting matrix ??

Title: Re: operators
Post by Obob on Nov 27^th, 2009, 10:12am

Correct, so the CONJUGATE transpose is

0 0 0 0 0 ....... 0 0
1 0 0 0 0 ........ 0 0
0 2 0 0 0 ........ 0 0
.
.
.
0 0 0 0 0 ...... 0 0
0 0 0 0 0 ...... m 0

(the conjugation being irrelevant in this example since all the entries are real)

Then this matrix represents the adjoint operator D*. In general, any time you have an orthonormal basis for a vector space and an operator E represented by a matrix in that basis, the adjoint operator is represented by the matrix E*. Depending on the definition of "adjoint" you are working with, there is something to prove here: the standard definition of an adjoint operator E* is that it is an operator such that <Ev,w> = <v,E*w> for all vectors v,w in your vector space. I'm telling you that E* is actually just the conjugate transpose, once you've written E down in terms of an orthonormal basis.

In this particular example, the operator D* takes 1 to x, x to 2x^2, x^2 to 3x^3, etc. So then if p(x) = sum ai x^i and q(x) = sum bi x^i are two polynomials,

<Dp,q> = sum i * a_{i-1} b_i as i goes from 1 to m
<p,D*q> = sum a_i * (i+1) b_(i+1) as i goes from 0 to m-1.

But these two sums are actually the same; just change the dummy variable i.

Title: Re: operators
Post by MonicaMath on Dec 1^st, 2009, 10:08am

So, in this case the general form of the adjoint will be

D*(q(x)) = sum (i+1) b_(i+1) x^i as i goes from 0 to m-1.

right !!?

Title: Re: operators
Post by Obob on Dec 1^st, 2009, 11:00am

on 12/01/09 at 10:08:16, MonicaMath wrote:

So, in this case the general form of the adjoint will be

D*(q(x)) = sum (i+1) b_(i+1) x^(i+1) as i goes from 0 to m-1.

right !!?

Almost. I fixed it in the quote.