# 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?

• Integrate by parts. Dec 18, 2020 at 17:50
• How to take the derivative of $\overline{g(t)}$.?
– PAMG
Dec 18, 2020 at 17:51
• It should be easy to show that $(\overline{g})' = \overline{g'}$, i.e. derivatives and complex conjugates commute. Dec 18, 2020 at 17:51
• 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 $T^* g$. Dec 18, 2020 at 17:54
• @Nate Eldredge I got one of the term $i(f(1)\overline{g(1)}-f(-1)\overline{g(-1)})$. How to handle this?
– PAMG
Dec 18, 2020 at 17:57

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.

• 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 $i\sqrt{6} u_1+2 u_2$. Dec 18, 2020 at 21:38
• @Christoph You're probably right Dec 18, 2020 at 21:46

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 $$T^*(1) = -3it+2,\quad D^*(t) = \frac 5 2i - \frac{15}2 i t^2 + 2t,\quad D^*(t^2) = -it+2t^2.$$