In , let T be defined by T(f) = f', where

Is T normal, self-adjoint, or neither?

In , define T by

Same problem here.

I don't know how to find T* from here. Please help, thanks.

Printable View

- Apr 11th 2008, 09:35 AMtttcomraderNormal, self-adjoint, or neither linear operator
In , let T be defined by T(f) = f', where

Is T normal, self-adjoint, or neither?

In , define T by

Same problem here.

I don't know how to find T* from here. Please help, thanks. - Apr 11th 2008, 11:30 AMOpalg
- Apr 11th 2008, 02:23 PMtttcomrader
Can I say ?

And for the second one, I have:

and

Then I'm bit lost, how I can find something that would fit ??? - Apr 12th 2008, 10:30 AMOpalg
For the second one, you're almost there. If the inner product is given by , and , then . But this is the same as (since ). So T is selfadjoint.

The first one is a good deal more complicated than I first thought, and I wonder if there is something wrong with the question. My original idea was to use integration by parts to express in terms of . But when you do the integration by parts, with the limits 0 and 1, you actually get . The awkward terms f(1)g(1)-f(0)g(0) prevent you from getting a neat expression for the adjoint of the differentiation operator. That seems to have the effect of making this question unreasonably difficult. - Apr 15th 2008, 06:24 AMtttcomrader
I asked my professor, and he said we should use this theorem:

T is normal if and only if there exist an orthonormal basis for V consisting of eigenvectors of T.

So, to find eigenvectors of T, I have So I need to find some polynomials f, are there any?

the standard ordered basis of V is

So the matrix

So the eigenvalue of T is 0,1.

Well, then any polynomial to the first power would satisfy T, right? - Apr 15th 2008, 12:04 PMOpalg
Just to be clear about this, I assume that means polynomials of degree at most 2 over the real numbers.

Interesting suggestion. I hadn't thought of doing it that way. There's a snag, though. That theorem works if the scalars are the complex numbers, but you have to be careful how to use it if the scalars are the reals. For example, the matrix is normal, but has no real eigenvalues or eigenvectors.

To use the theorem here, you must first show that the complex space does not have a base consisting of eigenvectors of T, and then deduce that since T is not normal as an operator on that space it must also fail to be normal when we restrict to real scalars.

An eigenvector would have to be a polynomial whose derivative is a multiple of itself. If f(x) is a constant, then its derivative is zero times itself; that's where the eigenvalue 0 fits in. The only other functions whose derivatives are multiples of themselves are exponentials (which are not polynomials, of course).