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Topic: operators (Read 2702 times) 

MonicaMath
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operators
« on: Nov 26^{th}, 2009, 11:11am » 
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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)


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Obob
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Re: operators
« Reply #1 on: Nov 26^{th}, 2009, 12:02pm » 
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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.


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MonicaMath
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Re: operators
« Reply #2 on: Nov 26^{th}, 2009, 11:28pm » 
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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 ??

« Last Edit: Nov 26^{th}, 2009, 11:29pm by MonicaMath » 
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Obob
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Re: operators
« Reply #3 on: Nov 27^{th}, 2009, 10:12am » 
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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_{i1} 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 m1. But these two sums are actually the same; just change the dummy variable i.


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MonicaMath
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Re: operators
« Reply #4 on: Dec 1^{st}, 2009, 10:08am » 
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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 m1. right !!?


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Obob
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Re: operators
« Reply #5 on: Dec 1^{st}, 2009, 11:00am » 
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on Dec 1^{st}, 2009, 10:08am, 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 m1. right !!? 
 Almost. I fixed it in the quote.


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