# Linear Algebra - Self-adjoint Operators

## Homework Statement

Make P2(R) into an inner-product space by defining <p, q> = $$\int_0^1p(x)q(x)dx$$. Define T in L(P2(R)) by T(a_0 + a_1*x + a_2*x2) = a_1*x.

(a) Show that T is not self-adjoint.
(b) The matrix of T with respect to the basis (1, x, x2) is

$$\left( \begin{array}{ccc} 0 & 0 & 0\\ 0 & 1 & 0\\ 0 & 0 & 0 \end{array} \right)$$

This matrix equals its conjugate transpose, even though T is not self-adjoint. Explain why this is not a contradiction.

## The Attempt at a Solution

To show that T is not self-adjoint, I need to show that T != T*. This amounts to explicitly finding what T* is. I know that <Tv, w> = <v, T*w>.

So we have <T(a_0 + a_1*x + a_2*x2), b_0 + b_1*x + b_2*x2> = <a_1*x, b_0 + b_1*x + b_2*x2> = a_1*x(b_0 + b_1*x + b_2*x2). Now I can't quite figure out where to go from here. I know this has to equal <a_0 + a_1*x + a_2*x2, T*(b_0 + b_1*x + b_2*x2)>. How can I find T*?

For part (b), I know that a if an operator is self-adjoint, its matrix equals its conjugate transpose, but not necessarily the other way around. This is a crappy way of putting it, though. Maybe I'll have a better explanation once I find T*. Thanks for any help! I wasn't paying attention to the problem and was calculating the inner-product wrong. It's obviously defined to be that integral between 0 and 1, not the dot product as I was doing before.

Can I just do this:

An operator is self-adjoint iff <Tv, w> = <v, Tw> for all v, w. So using the defined inner product, we have <T(a_0 + a_1*x + a_2*x2), b_0 + b_1*x + b_2*x2> = (1/2)*a_1*b_0 + (1/3)a_1*b_1 + (1/4)*a_1*b_2.

But we also have <a_0 + a_1*x + a_2*x2, T(b_0 + b_1*x + b_2*x2)> = (1/2)*a_0*b_1 + (1/3)a_1*b_1 + (1/4)*a_2*b_1.

Clearly <Tv, w> != <v, Tw>, so T is not self-adjoint. Is this correct?

Dick
Homework Helper
That's correct. You could even pick an easier example. <1,T(x)>=<1,x>!=0 and <T(1),x>=<0,x>=0. Not self adjoint.

That's correct. You could even pick an easier example. <1,T(x)>=<1,x>!=0 and <T(1),x>=<0,x>=0. Not self adjoint.

Cool. Thanks for the confirmation. Now I'm trying to figure out why I can say that even though my matrix equals its conjugate transpose, it is not a contradiction. I'm thinking it has to do with how sparse the matrix is, specifically on the diagonal, but I'm not sure. Anyone? Thanks!

Hurkyl
Staff Emeritus
Gold Member
I know that a if an operator is self-adjoint, its matrix equals its conjugate transpose,
Why do you think that?

Dick
I see now. Thanks for the clarification. 