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Confused about choice of vector in a proof.

  1. Oct 19, 2013 #1
    This is probably going to be a very simple question, i just need justification for a seemingly simple step in a proof.

    The statement is as follows:

    An endomorphism [itex]T[/itex] of an inner product space is [itex]{0}[/itex] if and only if [itex]\langle b|T|a\rangle = 0[/itex] for all [itex]|a\rangle[/itex] and [itex]|b\rangle[/itex].

    Now it is obvious if [itex]T[/itex] is [itex]0[/itex] then [itex]\langle b|T|a\rangle = 0[/itex]

    For the converse proof if [itex]\langle b|T|a\rangle = 0[/itex] for all [itex]|a\rangle[/itex] and [itex]b\rangle[/itex] then [itex]T = 0[/itex], it starts by choosing [itex]|b\rangle = T|a\rangle[/itex].

    Why is this valid, i guess a very naive reasoning would be doesn't this only prove it for the case that [itex]|b\rangle = T|a\rangle[/itex].
  2. jcsd
  3. Oct 20, 2013 #2


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    In the converse we're already given that [itex]\langle b|T|a\rangle = 0[/itex] for all [itex]|a\rangle[/itex] and [itex]b\rangle[/itex]. We don't have to prove anything for all ##|b\rangle##. We just consider ##T|a\rangle## and find that its norm must be zero for all ##|a\rangle##. Therefore ##T|a\rangle## is the zero vector for all ##|a\rangle##, hence ##T## is the zero map.
  4. Oct 21, 2013 #3


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    If the assumption in a proof is a "for all x in X" statement, you are always allowed to set x to any value you like. Often we are quite happy to take any allowed value ("let ##x \in X## arbitrary") as long as we know that it satisfies some condition, but you can always use that since the statement holds for all x in X, it holds for some particular convenient value.

    Of course, if you are trying to prove a statement of the form "for all x in X", it is not sufficient to prove the statement for a single particular x. For example, you wrote
    Actually what you meant is
    It would have been wrong to write
    Since that would not have been a statement about all |a⟩ and ⟨b|.
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