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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?

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  • $\begingroup$ Integrate by parts. $\endgroup$ Dec 18, 2020 at 17:50
  • $\begingroup$ How to take the derivative of $\overline{g(t)}$.? $\endgroup$
    – PAMG
    Dec 18, 2020 at 17:51
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    $\begingroup$ It should be easy to show that $(\overline{g})' = \overline{g'}$, i.e. derivatives and complex conjugates commute. $\endgroup$ Dec 18, 2020 at 17:51
  • $\begingroup$ The goal is to eventually rewrite it as $\int_{-1}^1 f(t) \left(\overline{\text{stuff involving $g$}}\right)\,dt$. Then the "stuff involving $g$" will be a formula for $T^* g$. $\endgroup$ Dec 18, 2020 at 17:54
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    $\begingroup$ @Nate Eldredge I got one of the term $i(f(1)\overline{g(1)}-f(-1)\overline{g(-1)})$. How to handle this? $\endgroup$
    – PAMG
    Dec 18, 2020 at 17:57

2 Answers 2

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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.

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  • $\begingroup$ I did a calculation without changing to an orthonormal basis and got a different result. I think you made a mistake calculating $[T]_{\mathcal B}$, since $T(u_2) = i\sqrt{3/2}+2 u_2 = i\sqrt{3} u_1 + 2 u_2$ instead of $i\sqrt{6} u_1+2 u_2$. $\endgroup$
    – Christoph
    Dec 18, 2020 at 21:38
  • $\begingroup$ @Christoph You're probably right $\endgroup$ Dec 18, 2020 at 21:46
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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 $$ T^*(1) = -3it+2,\quad D^*(t) = \frac 5 2i - \frac{15}2 i t^2 + 2t,\quad D^*(t^2) = -it+2t^2. $$

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