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Topic: operators (Read 2703 times) |
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MonicaMath
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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 26th, 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 26th, 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
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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 ??
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| « Last Edit: Nov 26th, 2009, 11:29pm by MonicaMath » |
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Obob
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Re: operators
« Reply #3 on: Nov 27th, 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
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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.
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MonicaMath
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Re: operators
« Reply #4 on: Dec 1st, 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 m-1.
right !!?
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Obob
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Re: operators
« Reply #5 on: Dec 1st, 2009, 11:00am » |
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on Dec 1st, 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 m-1.
right !!? |
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Almost. I fixed it in the quote.
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