Adjoint of the linear transformation over a vector space over complex numbers. Consider the vector space $V:=P_2(\mathbb{C})$ consisting of polynomials of degree at most $2$ with complex coefficients together with the following inner product $$\langle f, g\rangle=\int_{-1}^1f(t)\overline{g(t)}\, dt$$
Evaluate the adjoint of the linear transformation $T:V\to V$ by $$T(f)=if'+2f.$$
My effort:
I know that $T^*$ is a linear transformation which satisfies
$$\langle Tf, g\rangle=\langle f, T^*g\rangle.$$
Above implies that $$i\int_{-1}^1f'(t)\overline{g(t)}\, dt+2\int_{-1}^1f(t)\overline{g(t)}\, dt=\langle f, T^*g\rangle.$$
Please tell how to proceed next?
 A: The approach suggested in the comments is great for some problems like this, especially where the space in question involves some kind of "zero boundary condition". However, I find that for this problem it is more practical to proceed as follows.
To begin, apply the Gram Schmidt process to $\{1,t,t^2\}$ to get an orthonormal basis for $P_2(\Bbb C)$ relative to this inner product. We end up with
$$
u_1(t) = \frac 1{\sqrt{2}}, \quad u_2(t) = \sqrt{\frac 32} \,t, \quad u_3(t) = k\cdot[3t^2 - 1] 
$$
for some (real) $k$ such that $\langle u_3,u_3\rangle = 1$.
Relative to the basis $\mathcal B = \{u_1,u_2,u_3\}$, we find that $T$ has the matrix
$$
[T]_{\mathcal B} = 
\pmatrix{2&i\sqrt{3}&0\\
0&2&i\sqrt{\frac 23}\cdot \frac 1{6k}\\
0&0&2}.
$$
The matrix of $T^*$ relative to this same basis is the conjugate-transpose, which is to say that
$$
M = [T^*]_{\mathcal B} = 
\pmatrix{2&0&0\\
-i\sqrt{3} & 2 & 0\\
0 & -i\sqrt{\frac 23}\cdot \frac 1{6k} & 2}.
$$
This matrix implicitly gives us a formula for $T^*$. In particular,
$$
[T^*p](t) = \sum_{j=0}^2 \left[\sum_{k=1}^3 M_{jk}\langle p,u_k\rangle\right] u_j(t).
$$
I see no nice way to interpret the operator that we get in this case in terms of derivatives and integrals.
A: It is enough to consider the transformation $D(f)=f'$, since $T=iD+2\operatorname{id}$ yields $T^*=-iD^*+2\operatorname{id}$.
Instead of constructing an orthonormal basis for $P_2(\mathbb C)$ first, we can do the computation in our favorite basis $B=(1,t,t^2)$ using the Gram matrix of the given inner product.
The Gram matrix of $\langle -,-\rangle$ with respect to $B$ is obtained from the integrals $\int_{-1}^1 t^k\,\mathrm dt$ for $k=0,1,2,3,4$ as
$$
G_B = (\langle t^i, t^j\rangle)_{i,j=0,1,2} =
\begin{pmatrix}
2 & 0 & 2/3 \\
0 & 2/3 & 0 \\
2/3 & 0 & 2/5
\end{pmatrix}.
$$
Denoting the coordinate vector of $f\in P_2(\mathbb C)$ with respect to $B$ by $[f]_B$ this describes our inner product as
$$
\langle f,g\rangle = [f]_B^T \ G_B \ \overline{[g]_B}.
$$
The matrix of $D$ with respect to $B$ is given by
$$
[D]_B =
\begin{pmatrix}
0 & 1 & 0 \\
0 & 0 & 2 \\
0 & 0 & 0
\end{pmatrix}.
$$
Now by comparing the descriptions
\begin{align*}
\langle Df,g\rangle &= [f]_B^T\ [D]_B^T \ G_B \ \overline{[g]_B} \qquad\text{and} \\
\langle f,D^*g\rangle &= [f]_B^T\ G_B \ \overline{[D^*]_B}\ \overline{[g]_B},
\end{align*}
one obtains the general formula for the matrix of the adjoint in non-orthonormal bases:
$$
[D^*]_B = \overline{G_B^{-1} [D]_B^T \ G_B} =
\begin{pmatrix}
0 & -5/2 & 0 \\
3 & 0 & 1 \\
0 & 15/2 & 0
\end{pmatrix}.
$$
Hence $D^*$ is given by
$$
D^*(1) = 3t,\quad D^*(t) = -\frac 5 2+ \frac{15}2 t^2,\quad D^*(t^2) = t
$$
which translates to $T^*$ as
$$
T^*(1) = -3it+2,\quad D^*(t) = \frac 5 2i - \frac{15}2 i t^2 + 2t,\quad D^*(t^2) = -it+2t^2.
$$
