# Normal, self-adjoint, or neither linear operator

• Apr 11th 2008, 08:35 AM
Normal, self-adjoint, or neither linear operator
In $P_{2}( \mathbb {R} )$, let T be defined by T(f) = f', where $ = \int _{0}^{1} f(t)g(t)dt$

Is T normal, self-adjoint, or neither?

In $M_{2x2} ( \mathbb {R} )$, define T by $T(A) = A^t$

Same problem here.

• Apr 11th 2008, 10:30 AM
Opalg
Quote:

In $P_{2}( \mathbb {R} )$, let T be defined by T(f) = f', where $ = \int _{0}^{1} f(t)g(t)dt$

Is T normal, self-adjoint, or neither?

In $M_{2x2} ( \mathbb {R} )$, define T by $T(A) = A^t$

Same problem here.

I don't know how to find T* from here.

The adjoint operator is given by $\langle T^*f,g\rangle = \langle f,Tg\rangle$. If Tg = g', and the inner product is given by integration over the unit interval, then $\langle f,Tg\rangle = \int _{0}^{1} f(t)g'(t)\,dt$. You want to write this in the form $\int _{0}^{1} ???g(t)\,dt$, where "???" will be T*(f).
• Apr 11th 2008, 01:23 PM
Can I say $T^*(f) = \frac {f(t)g(t)}{g'(t)}$?

And for the second one, I have:

$ = $

$==tr(B^t,A^t)$

and $==tr(???^t,A)$

Then I'm bit lost, how I can find something that would fit ???
• Apr 12th 2008, 09:30 AM
Opalg
For the second one, you're almost there. If the inner product is given by $\langle A,B\rangle = \text{tr}(B^{\textsc T}A)$, and $T(A) = A^{\textsc T}$, then $\langle T(A),B\rangle = \text{tr}(B^{\textsc T}A^{\textsc T})$. But this is the same as $\langle A,T(B)\rangle$ (since $\text{tr}(M) = \text{tr}(M^{\textsc T})$). 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 $\textstyle\int fg'$ in terms of $\textstyle\int f'g$. But when you do the integration by parts, with the limits 0 and 1, you actually get $\langle f',g\rangle = f(1)g(1)-f(0)g(0) - \langle f,g'\rangle$. 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, 05:24 AM
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 $T(f)=f'= \lambda f$ So I need to find some polynomials f, are there any?

the standard ordered basis of V is $\{ 1,x,x^2 \}$

So the matrix $[T]_{ \beta } = \begin{pmatrix} 0 && 0 && 0 \\ 0 && 1 && 2 \\ 0 && 0 && 0 \end{pmatrix}$

So the eigenvalue of T is 0,1.

Well, then any polynomial to the first power would satisfy T, right?
• Apr 15th 2008, 11:04 AM
Opalg
Quote:

In $P_{2}( \mathbb {R} )$, let T be defined by T(f) = f', where $ = \int _{0}^{1} f(t)g(t)dt$

Is T normal, self-adjoint, or neither?

Just to be clear about this, I assume that $P_{2}( \mathbb {R} )$ means polynomials of degree at most 2 over the real numbers.

Quote:

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.

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 $\begin{bmatrix}0&1\\-1&0\end{bmatrix}$ is normal, but has no real eigenvalues or eigenvectors.

To use the theorem here, you must first show that the complex space $P_{2}( \mathbb {C} )$ 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.

Quote:

So, to find eigenvectors of T, I have $T(f)=f'= \lambda f$ So I need to find some polynomials f, are there any?
the standard ordered basis of V is $\{ 1,x,x^2 \}$
So the matrix $[T]_{ \beta } = \begin{pmatrix} 0 & 0 & 0 \\ 0 & 1 & 2 \\ 0 & 0 & 0 \end{pmatrix}$ Not quite; the 1 should be at the bottom of the middle column.