Finding the adjoint of operator on P_1

Question

The question asks to find the adjoint of the operator T on P$_1$([0,1]), which is the space of polynomials with degree no greater than 1 over the field [0,1], which contains all numbers between 0 and 1, inclusive. Specifically, the operator T is defined as: $$p → \int_0^1 x^{k} p(x) dx$$ What I've tried so far is I tried solving T(f), where f ∈ P$_1$([0,1]), and then plugging it into the inner product on the space of functions, which is <f,g> = $\int_0^1 f(x) g(x) dx$, and calculating that with g, where g ∈ P$_1$([0,1]), but then that doesn't really give me anything, just a complicated constant and the integral of g(x). Someone else I asked suggested I try using the inner product immediately, without solving for T(f), but I'm not sure how that would help any more than what I already did.

Any help would be appreciated, thank you in advance. I'm a first-year undergrad student by the way, if that affects any answers.

Answer 1

According to the definition $T(a,b)=\int_{0}^{1}x^{k}(ax+b)dx=\dfrac{a}{k+2}+\dfrac{b}{k+1}$=$\begin{bmatrix} \dfrac{1}{k+2}&\dfrac{1}{k+1} \\ \end{bmatrix}\begin{bmatrix} a\\b \end{bmatrix}$.

Then $T^{\star}:\mathbb{R} \to \mathbb{R}^{2}$ is

$T^{\star}(y)=\begin{bmatrix} \dfrac{1}{k+2}\\\dfrac{1}{k+1} \end{bmatrix}y$

Answer 2

I believe what that someone else was referring to was writing $\langle Tf,g\rangle$ as a double integral and using Fubini's theorem, but that doesn't actually give you an operator in $P_1[0,1]$.

Instead, we use what George outlined: if $ax+b$ corresponds to $\begin{pmatrix} a \\ b\end{pmatrix}$, then we see that the inner product $\langle \cdot, \cdot\rangle$ corresponds to the matrix $$P = \begin{pmatrix} \frac{1}{3} & \frac{1}{2} \\ \frac{1}{2} & 1\end{pmatrix}.$$

Likewise, as George said, $$T \begin{pmatrix}a \\ b\end{pmatrix} = \begin{pmatrix} 0 \\ \frac{a}{k+2} + \frac{b}{k+1}\end{pmatrix} \implies T = \begin{pmatrix} 0 & 0 \\ \frac{1}{k+2} & \frac{1}{k+1} \end{pmatrix}.$$

Now, suppose $$f = \begin{pmatrix}a \\ b\end{pmatrix},\quad g = \begin{pmatrix} c\\d\end{pmatrix}.$$ In matrix notation, we need to find a matrix $S$ such that $$(Tf)^T Pg = f^TP(Sg).$$ For this to hold true, we need $T^TP = PS,$ or $$S = P^{-1}T^TP.$$ For $P$ the identity matrix, this would mean $S = T^T$, but as $P$ isn't the identity matrix, we would need to take matrix inverses.