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Hermitian adjoint

  1. Jul 17, 2016 #1
    1. The problem statement, all variables and given/known data

    Be V the set ##\{f \in \mathbb{R}[X]| 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^*\,\,\,##.

    2. Relevant equations

    3. The attempt at a solution
    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?
    For b), i dont find a starting point.
     
  2. jcsd
  3. Jul 17, 2016 #2

    andrewkirk

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    Can't we approach (b) the same way as (a)?
    For (b), the defining equation is
    $$\langle\Phi(f),s\rangle=\langle f,\Phi^*(s)\rangle$$
    What happens if we expand that using the definitions given?

    Your working for (a) looks broadly correct. To check that something has not gone wrong, like a missed sign, plug a polynomial ##f(x)=a_0+a_1x+a_2x^2## into it and see if the equality of the two inner products holds.
     
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