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Hermitian adjoint operators (simple "proofs")

  1. Oct 25, 2015 #1
    1. The problem statement, all variables and given/known data
    I'm having some trouble with questions asking me to "show" or "prove" instead of computing an answer so I'm looking for some input if I'm actually doing what I'm supposed to or not (and for the last one I don't know where to get started really.)
    1. Show that ##T^*## is linear.
    2. Show that ##(T^*)^* = T##.
    3. Show that ##\langle v,Tw\rangle = \langle T^*v,w\rangle##.
    4. Show that ##(ST)^* = T^*S^*##.

    I realise this is quite a lot so even taking a look at one of the above would be very kind!

    2. Relevant equations
    The adjoint to the linear operator ##T:V \to W## is the mapping ##T^*:W \to V## that is defined by ##\langle Tv,w \rangle = \langle v,T^*w\rangle## for all ##v\in V## and ##w \in W##.

    3. The attempt at a solution
    1. Using the definition for adjoint and the scalar product we have
    ##\langle v, T^*(w+u)\rangle = \langle Tv,w+u \rangle = \langle Tv,w\rangle + \langle Tv,u\rangle = \langle v,T^*w\rangle + \langle v,T^*u\rangle = \langle v,T^*w +T^* u \rangle##.
    And for ##\langle v, T*(\alpha w)\rangle = \langle Tv,\alpha w\rangle = \bar \alpha \langle Tv,w\rangle = \bar \alpha \langle v, T^* w \rangle = \langle v, \alpha T^*w\rangle##.
    Since ##v## is any vector in ##V## we have that ##T^*(v+w) = T^*v+T^*w## and ##T^*(\alpha v) = \alpha T^*v## are the only possibilities and that should show the linearity.

    2. From the definition and using the conjugate symmetry of the scalar product we have
    ##\langle T^*v,w \rangle = \langle v, (T^*)^*w\rangle = \overline{ \langle (T^*)^*,v \rangle }##
    But we also have
    ##\langle T^*v,w \rangle = \overline{\langle w, T^*v\rangle}= \overline{\langle Tw,v\rangle } ##. Again since ##v## is any vector in ##V## we have ##(T^*)^* = T##.

    3. As I understand the question this means either that the operators switches roles i.e. ##T: W\to V## and ##T^*:V\to W##. Is this true or I'm supposed to show this when the operators don't even operate on the same space?? If it's as I imagine, doesn't this follow immediately from (2)?

    4. This is the one I'm not sure how to do. Writing up the definition I have and using (2)
    ##\langle STv , w\rangle = \langle v, (ST)^*w\rangle = \langle (S^*T^*)^*v,w \rangle## and I don't know how to push ahead.
     
  2. jcsd
  3. Oct 25, 2015 #2

    PeroK

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    Your answers to 1-2 look good to me. For 3, there's no law that says that T must act on V into W. An arbitrary linear operator T can just as well act on W into V. And, for 3, T is just such an operator.

    For 4, hint ##Tv## is a vector.
     
  4. Oct 25, 2015 #3
    Thank you for going over the exercises! So for (3) is ##T## and operator ##T:U\to U## where ##U = V\cup W##? Or did I misunderstand you?
    If this is true then ##\langle v, Tw \rangle = \overline{\langle Tw,v \rangle } = \overline{\langle w,T^*v \rangle} = \langle T^*v,w\rangle##.

    That hint made it a whole lot of easier ##\langle v,(ST)^*w \rangle = \langle STv,w\rangle = \langle Tv,S^*w \rangle = \langle v,T^*S^*w \rangle##.
     
  5. Oct 25, 2015 #4

    PeroK

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    In 3, you can infer that ##T: W \rightarrow V##
     
  6. Oct 25, 2015 #5
    Alright so it's enough that ##T: W \rightarrow V## with ##T:U\to U## being a special case. But other than that it's correct?
     
  7. Oct 25, 2015 #6

    PeroK

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    Yes. In general, the linear operator may map one inner product space to another; or, the two spaces could be the same.
     
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