Let $A\in\mathcal L\left(\mathcal P_2\right)$ be defined as $(Ap)(t)=p'(t)$. Find the adjoint operator $A^*$ if the inner product on $\mathcal P_2$ is defined as: $\langle p,q\rangle=p(0)q(0)+p(1)q(1)+p(2)q(2),\quad\forall p,q\in\mathcal P_2$.
Note: $\mathcal P_2$ denotes the vector space of polynomials of degree $\deg\le 2$ with real coefficients.
My attempt:
Adjoint operator $A^*\in\mathcal L\left(P_2\right)$ satisfies $\langle Ap,q\rangle=\langle p,A^*q\rangle$. Let $p(t)=at^2+bt+c$ and $q(t)=dt^2+et+f$, where $a,b,c,d,e,f\in\Bbb R$. Then $(Ap)(t)=2at+b$ and $$\begin{aligned}\langle Ap, q\rangle &= p'(0)q(0)+p'(1)q(1)+p'(2)q(2)\\&=b\cdot f+(2a+b)\cdot(d+e+f)+(4a+b)\cdot(4d+2e+f)\\&=\langle p, A^*q\rangle,\end{aligned}$$
but I couldn't figure out how to transform this expression into one of the form: $$p(0)(A^*q)(0)+p(1)(A^*q)(1)+p(2)(A^*q)(2). $$
I know the matrix representation in the monomial basis $e=\{1,t,t^2\}$ would be $[A^*]_e^e=\begin{bmatrix}0&0&0\\0&0&0\\0&0&1\end{bmatrix}$. However, I'm not sure how to deal with the inner product defined this way.
Edit:
If we rewrite $p'(t)q(t)=(Apq)(t)-p(t)(Aq)(t)$, could we rewrite $\sum_{t=0}^2(Apq)(t)$ as $\langle p, Bq\rangle$, where $B$ is the identity, so that $A^*=(B-A)$?
May I ask for help how to continue with his task?
Thank you very much!
Best Answer
Hints. Let $E$ be the ordered basis $\{x^2,x,1\}$. Suppose you can find an orthonormal basis $B=\{f,g,h\}$ of $\mathcal P_2$ with respect to the given inner product. (Consider some appropriate Lagrange interpolation polynomials!) Then \begin{align} [A^\ast]_B^B &=\left([A]_B^B\right)^\ast\\ &=\left([I]_E^B\,[A]_E^E\,[I]_B^E\right)^\ast\\ &=\left([I]_B^E\right)^\ast \left([A]_E^E\right)^\ast\ \left([I]_E^B\right)^\ast\\ &=\left([I]_B^E\right)^\ast \left([A]_E^E\right)^\ast\ \left(\left([I]_B^E\right)^{-1}\right)^\ast \end{align} where $I$ denotes the identity map. If you want to find the matrix of $A^\ast$ in the basis $E$, you may perform a step further: \begin{align} [A^\ast]_E^E &=[I]_B^E\ [A^\ast]_B^B\ [I]_E^B\\ &=[I]_B^E\,\left([I]_B^E\right)^\ast\ \left([A]_E^E\right)^\ast \ \left(\left([I]_B^E\right)^{-1}\right)^\ast\,\left([I]_B^E\right)^{-1}\\ &=[I]_B^E\,\left([I]_B^E\right)^\ast\ \left([A]_E^E\right)^\ast \ \left([I]_B^E\,\left([I]_B^E\right)^\ast\right)^{-1}. \end{align}