# Thread: Normal, self-adjoint, or neither linear operator

1. ## 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.

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

2. Originally Posted by tttcomrader
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).

3. 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 ???

4. 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.

5. 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?

6. Originally Posted by tttcomrader
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.

Originally Posted by tttcomrader
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.

Originally Posted by tttcomrader
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.

So the eigenvalue of T is 0,1. No, 0 is the only eigenvalue.

Well, then any polynomial to the first power would satisfy T, right?
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).