Hermitian adjoint

Question

I'm trying to solve this task, but I'm not sure, if my solution for a) is correct. For b), i dont find a starting point. Did someone have an idea how to solve this?

Thanks in advance.

Be $V$ the set $\{f \in \mathbb{R}[X]\mid \operatorname{deg}\,f \leq 2 \}$. This becomes to an euclidic vector space through the inner product $\langle f,g\rangle:=\sum_{i=-1}^1f(i)g(i)$ . The same goes for $\mathbb{R}$ with the inner product $\langle r,s\rangle :=rs$.

a) For $j:\mathbb{R}\to V,r\mapsto rX$, calculate the hermitian adjoint $j^*$.

b) Be $\Phi :V \to \mathbb{R}$ the linear map $\sum_{i=0}^2a_iX^i \mapsto \sum_{i=0}^2a_i \,\,\,$. Calculate the hermitian adjoint $\Phi^*$.

For a) I have the follwowing solution:

$\langle f,j(s) \rangle_V = \langle j^*(f), s \rangle_{\mathbb{R}}$

$\Rightarrow \sum_{i=-1}^1f(i) \cdot (j(s))(i)=j^*(f) \cdot s$

$\Rightarrow f(-1)\cdot -s+f(0)\cdot 0s+f(1)\cdot s = j^*(f) \cdot s$

$\Rightarrow j^*(f)=f(1)-f(-1)$

Is this solution correct?

Did someone have any ideas for b)?

Answer 1

Your solution for a) is correct, except that you ought to write $\iff$ instead of $\implies$, as we mostly care for the $\Leftarrow$ direction.

Proceeding the same way for b):

Let $f(X) = a_0 + a_1X + a_2X^2$. Then

$\begin{align} & \langle r, \Phi(f)\rangle_\mathbb{R} = \langle \Phi^*(r),f\rangle_V \\ \iff & r\Phi(f) = \Phi^*(r)(-1)\cdot f(-1) + \Phi^*(r)(0) \cdot f(0) + \Phi^*(r)(1) \cdot f(1) \\ \iff & r(a_0+a_1+a_2) = \Phi^*(r)(-1)\cdot (a_0 - a_1 + a_2) + \Phi^*(r)(0) \cdot a_0 + \Phi^*(r)(1) \cdot (a_0+a_1+a_2). \\ \end{align}$

But this must hold for arbitrary $f$, and so we may choose particular coefficients $a_i$ to discover what $\Phi^*(r)$ must be.

• $a_0 = 0$, $a_1 = a_2 = 1$ yields $2r = 2\Phi^*(r)(1)$, i.e. $\Phi^*(r)(1) = r$.
• $a_0 = a_1 = 0$, $a_2 = 1$ yields $r = \Phi^*(r)(-1) +r$, i.e. $\Phi^*(r)(-1) = 0$.
• $a_0 = 1$, $a_1=a_2=0$ yields $r = \Phi^*(r)(0) + r$, i.e., $\Phi^*(r)(0) = 0$.

From this information we conclude that $\Phi^*(r)(X) = (rX + rX^2)/2$ is the adjoint of $\Phi$.