adjoint of an operator I have a very simple question
Let $V$ be the real (finite-dimensional) inner product space of polynomials of degree at most $2$, with inner product $(p,q):= \int_0^1 p(x)q(x) \, dx$.
Let $T$ be the linear operator that maps $p_0+p_1x+p_2x^2$ to $p_1x$. The question is how to find the adjoint of $T$.
I have one way to solve the problem: namely find an orthonormal basis of $V$, find the matrix
associated to $T$ under this basis and then take its transpose to get the matrix of $T^*$ under this basis.
Is there any quicker/smarter way?
Thanks a lot in advance
Jenny
 A: It is funny to note that integration by parts
$$(Tp,q)=\int_0^{1}p_1 x q(x)dx=p_1\left[g(1)-\int_{0}^1 g(x)dx \right],$$
provides us with little information as $g(x)$, which satisfies $g'(x)=q(x)$ is not an element of $V$ if the coefficient $q_2$ of the polynomial 
$$q(x)=q_0+q_1x +q_2x^2$$
is different from $0$.
Let us try another method. By definition, $T^{\dagger}(q)\in V$, where
$$T^{\dagger}(q):=\alpha_0+\alpha_1 x+\alpha_2 x^2,$$
with $\alpha_i=\alpha_i(q_0,q_1,q_2)$ for all $i$ and 
$$(T(p),q)=(p,T^{\dagger}(q)),$$
for all $p,q\in V$. 
We select different $p$'s and try to arrive at information for the functions $\alpha_i$.


*

*$p(x)=1$


In this case we have $0=(T(1),q)=(1,T^{\dagger}(q))$, or
$$\int_0^{1} T^{\dagger}(q)dx=\alpha_0+\frac{\alpha_1}{2}+\frac{\alpha_2}{3}=0.$$


*

*$p(x)=x$


In this case we have $(T(x),q)=(x,T^{\dagger}(q))$, or
$$\int_0^{1} x q(x) dx=\int_0^{1} xT^{\dagger}(q)dx$$
which is equivalent to $\delta=\frac{\alpha_0}{2}+\frac{\alpha_1}{3}+\frac{\alpha_2}{4},$
where
$$\delta=\frac{q_0}{2}+\frac{q_1}{3}+\frac{q_2}{4}. $$


*

*$p(x)=x^2$


In this case we have $0=(T(x^2),q)=(x^2,T^{\dagger}(q))$, or
$$\int_0^{1} x^2T^{\dagger}(q)dx=\frac{\alpha_0}{3}+\frac{\alpha_1}{4}+\frac{\alpha_2}{5}=0.$$
In summary, we arrive at the system of equations
$$\alpha_0+\frac{\alpha_1}{2}+\frac{\alpha_2}{3}=0 $$
$$\frac{\alpha_0}{2}+\frac{\alpha_1}{3}+\frac{\alpha_2}{4}=\delta $$
$$\frac{\alpha_0}{3}+\frac{\alpha_1}{4}+\frac{\alpha_2}{5}=0 $$
which admits the solution 
$$\alpha_0=-36\delta $$
$$\alpha_1=192\delta $$
$$\alpha_2=-180\delta. $$
I do not know if this method is quicker than orthonormalization, but I hope it helps.
