Calculate adjoint operator inner product

Question

Consider the inner product of $P_1(\mathbb R)$ given by $$\langle p(x),q(x)\rangle = \int_0^1p(x)q(x)x^2 dx\quad\text{(note the factor $x^2$ in the integrand)}$$ Let $T\colon P_1(\mathbb R)\to P_1(\mathbb R)$ be given by $T(p(x))= p(x)+p'(x)$. Determine $T^*(a+bx)$ where $a,b\in\mathbb R$ are arbitrary scalars.

This is what I've done, don't think the answer is right though so need help with what I'm doing wrong and the right way to think:

Adjoint mapping $T^*$ of $T$ is the linear operator satisfying the following property for all $p(x), q(x) \in P_1(\mathbb R)$:

$$\langle T(p(x)), q(x)\rangle = \langle p(x), T^*(q(x))\rangle$$

First, we calculate $T(p(x))$: $$T(p(x)) = p(x) + p'(x)$$

Now we can use this to find $T^*(a+bx)$ for arbitrary scalars $a$ and $b$. We use the definition of the inner product:

\begin{align*} \langle T(a+bx), q(x)\rangle &= \langle (a+bx) + (a+bx)', q(x)\rangle \\ &= \rangle (a+bx) + b, q(x)\rangle \end{align*}

Now we need to find $T^*(q(x))$. Since $T^*$ is a linear operator, we can write $T^*(q(x))$ as $T^*(q(x)) = c + dx$, where $c$ and $d$ are unknown constants.

So now we have:

$$\langle T^*(q(x)), p(x)\rangle = \langle c + dx, p(x)\rangle$$

To get $T^*(q(x))$ from this, we need to compare the two sides of the equation and determine the constants $c$ and $d$.

To do so, we can use the properties of the given inner product:

\begin{align*} \langle c + dx, p(x)\rangle &= \int_0^1 (c+dx)p(x)x^2\,dx \end{align*}

To match the two sides of the equation above, one must:

$$c + dx = (a+bx) + b$$

This gives us two equations:

$$c = a + b$$ $$d = b$$

So $T^*(q(x)) = a + b + x$. Therefore is:

$$T^*(a+bx) = a + b + x$$

Answer 1

To begin with, let's start considering that $p(x) = a + bx$ is equivalent to the coordinates $(a,b)$ (linear combination of the canonical base). Since $T^* : P_1[x] \rightarrow P_1[x]$, $T^*(p(x))$ has two coordinate functions, $T^*(p(x)) = (T^*_1(p(x)),T^*_2(p(x)))$. Let's start with the element $1$, which has coordinates $(1,0). T((1,0)) = (1,0).$ Let $T^*(p(x)) = (c,d).$ $\langle (1,0),(c,d) \rangle = \langle (1,0),T^*(c,d)\rangle $, by definition of adjoint operator. The left element of the last equality is $\frac{c}{3} = \frac{1}{3}T^*_1(c,d)$, and therefore the first coordinate function of the adjoint is $ T^*_1(c + dx) = c$, this is, the constant term remains the same.
The next step would be using $x$ instead of $1$. This requires some additional calculations since there are more terms, but it is the same, using now the fact that you know that the first coordinate function of the adjoint is $c$. Of course, this is using the fact that $\langle (a + bx),(c + dx)\rangle = \langle (a,b),(c,d) \rangle = \frac{bd}{5} + \frac{ab+bc}{4} + \frac{ac}{3}$. I would appreciate if you gave upvote if this was useful

Answer 2

You need to use the definition of the adjoint here to your advantage. First note that in the canonical basis $\{1,x\}\to \{(1,0)^T,(0,1)^T\}$ we can represent $T$ as a matrix:

$$T=\begin{pmatrix}1&1\\1&0\end{pmatrix}$$

Since the adjoint also needs to be defined in the same space, it will also be a $2\times 2$ matrix whose most general form is

$$T^*=\begin{pmatrix}\alpha&\gamma\\\beta&\delta\end{pmatrix}$$

We can write a matrix equation for $T^*$ by looking at the inner products with the basis and using the definition of the adjoint to calculate them in two ways. To wit,

$$\langle T^*(1),1\rangle=\alpha\langle1,1\rangle+\beta\langle x,1\rangle=\langle1,T(1)\rangle$$ $$\langle T^*(1),x\rangle=\alpha\langle1,x\rangle+\beta\langle x,x\rangle=\langle1,T(x)\rangle$$ $$\langle T^*(x),1\rangle=\gamma\langle 1,1\rangle+\delta\langle x,1\rangle=\langle x,T(1)\rangle$$ $$\langle T^*(x),x\rangle=\gamma\langle 1,x\rangle+\delta\langle x,x\rangle=\langle x,T(x)\rangle$$

These equations can be recast in matrix form

$$\begin{pmatrix}\langle 1,1\rangle&\langle x,1\rangle\\\langle 1,x\rangle&\langle x,x\rangle\end{pmatrix}\begin{pmatrix}\alpha&\gamma\\\beta&\delta\end{pmatrix}=\begin{pmatrix}\langle 1,T(1)\rangle&\langle x,T(1)\rangle\\\langle 1,T(x)\rangle&\langle x,T(x)\rangle\end{pmatrix}$$

and now one can invert the inner product matrix in the front to obtain values of $\alpha, \beta,\gamma, \delta$.

For finite-dimensional linear spaces with an inner product this procedure can be easily generalized. In fact, the following matrix equation holds

$$T^{*}=GT^{\dagger} G^{-1}$$

where the dagger denotes hermitian conjugation, and $G_{ij}=\langle e_i, e_j\rangle$ is the Gram matrix. When the basis is orthonormal with respect to the inner product, $G$ is the identity, and hence the well-known fact that the adjoint is the hermitian conjugate follows.

Can you proceed from here?

$\alpha=-19,\gamma=-15, \beta=80/3, \delta=21$