Specific non self-adjoint operator for $\mathcal{P}^2$

Question

Given the inner product space $\mathcal{P}^2$ with inner product defined as $(<p,q>) =\int_0^1 p(x)q(x) dx$ and an operator $$T(a_0 + a_1x + a_2x^2) = a_1x$$I need to show that the operator is not self-adjoint, but I'm finding that

ANSWERED: ($$(<Tv,v>) = (<v,Tv>) = \int_0^1 a_0 + a_1^2x^2 + a_2x^2 dx = a_0 + \frac{1}{3}(a_1^2 + a_2)$$ What am I doing wrong?)

I was showing that $<Tv,v> = <v,Tv>$ instead of $<v,Tw>=<w,Tv>$.

Also, how come even though the conjugate transpose of the matrix of T is T the operator isn't self-adjoint?

Thanks in advance :)

Answer 1

You can consider $v=1$ and $w=x$: $$ \langle T(v),w\rangle=0,\quad \langle v,T(w)\rangle=\int_0^1x \ dx=\frac12. $$

Although the matrix for $T$ in the ordered basis $(1,x,x^2)$ is given by $$ \begin{pmatrix} 0&0&0\\ 0&1&0\\ 0&0&0 \end{pmatrix}, $$ which is a symmetric matrix, $T$ is not self-adjoint with respect to the inner product space endowed with your inner product.

If you consider the standard inner product instead, the $T$ is self-adjoint:

$$ ( a_0+a_1x+a_2x^2,b_0+b_1x+b_2x^2):=a_0b_0+a_1b_1+a_2b_2. $$ Then you have $(Tv,w)=(v,Tw)$ for all $v,w\in\mathcal{P}^2$.

Denote by $ (\cdot ,\cdot )$ the standard inner product on $\mathbb{R}^n$. The real $n$-by-$n$ matrix $A$ is symmetric if and only if $$ {\displaystyle (Ax,y)=(x,Ay)\quad \forall x,y\in {\mathbb {R} }^{n}.} $$