- #1

steelphantom

- 159

- 0

## Homework Statement

Make P

_{2}(R) into an inner-product space by defining <p, q> = [tex]\int_0^1p(x)q(x)dx[/tex]. Define T in L(P

_{2}(R)) by T(a_0 + a_1*x + a_2*x

^{2}) = a_1*x.

(a) Show that T is not self-adjoint.

(b) The matrix of T with respect to the basis (1, x, x

^{2}) is

[tex]\left(

\begin{array}{ccc}

0 & 0 & 0\\

0 & 1 & 0\\

0 & 0 & 0

\end{array}

\right)[/tex]

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

## Homework Equations

## 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*x

^{2}), b_0 + b_1*x + b_2*x

^{2}> = <a_1*x, b_0 + b_1*x + b_2*x

^{2}> = a_1*x(b_0 + b_1*x + b_2*x

^{2}). 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*x

^{2}, T*(b_0 + b_1*x + b_2*x

^{2})>. 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!