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Proof involving complex conjugates and Matrices

  1. Sep 19, 2015 #1

    RJLiberator

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    1. The problem statement, all variables and given/known data

    Show that (A+B)*=A*+B*

    2. Relevant equations
    I think I am missing a property to prove this.

    3. The attempt at a solution
    This should be easier then I am making it out to be. But I seem to be missing one key property to do this.

    A*+B* is just A(ij)*+B(ij)* = Right hand side
    Left hand side is just (A+B)(ij)*

    What am I missing here to make this concrete?

    I've looked around the web and found things such as A(ij)* = bar(A(ji))
     
  2. jcsd
  3. Sep 19, 2015 #2
    Well you know how each element of the matrix ##A+B## looks like right? Namely its ##a_{ij}+b_{ij}## and you know that the conjugate of a sum of numbers is the sum of the conjugates.

    Which makes you get that each element of the new matrix ##(A+B)^*## is ##a_{ij}^*+b_{ij}^*## which should remind you of something important
     
  4. Sep 19, 2015 #3

    Ray Vickson

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    What is your definition of A*? Is it the element-by-element complex conjugate of A, or is it the element-by-element complex conjugate of the transpose of A? I have seen the same notation A* used in both ways, so you really do need to clear that up.
     
  5. Sep 19, 2015 #4
    You could prove it by definition:
    Let ##H## be a Hilbert space and let ## A : H \rightarrow H ## be a bounded linear operator. Then, ## A^{*} ##, the adjoint of ##A##, is defined by the following equation: $$<Ax,y>=<x,A^{*}y> for \: all \: x,y \in H$$
    Thus, you have to prove that: $$<x,(A+B)^{*}y>=<x,(A^{*}+B^{*})y> for \: all \: x,y \in H$$
     
  6. Sep 19, 2015 #5

    RJLiberator

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    Is this what we are trying to prove? Essentially?

    If I say Left hand side: (Aij+Bij)*=Aij*+Bij*
    and so this equals the right hand side Aij* + Bij*

    Isn't that what we are just trying to prove? That would be a 2 line proof for it. That simple?
     
  7. Sep 19, 2015 #6

    RJLiberator

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    Most simple form used here: Element-by-element complex conjugate of A.
     
  8. Sep 19, 2015 #7
    Yes it would look like that. Just think of what happens to each element. It is a simple proof after all.
     
  9. Sep 19, 2015 #8

    RJLiberator

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    Okay, so
    A*+B*=Aij*+Bij* = Right hand side
    (A+B)*=Aij*+Bij* = Left Hand side due to the conjugate of a sum of numbers is the sum of the conjugates.

    0_o
     
  10. Sep 19, 2015 #9
  11. Sep 19, 2015 #10

    RJLiberator

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    Thanks for your help David.
     
  12. Sep 19, 2015 #11
    No problem, glad to help.
     
  13. Sep 19, 2015 #12
    It`s important to keep in mind that the operator ##^{*}## also involves taking the transpose of the matrix (if you choose the appropriate basis)
     
  14. Sep 19, 2015 #13

    vela

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    Your confusion probably arises because you're not paying enough attention to details, which is suggested by what you've written. For example, A*+B* isn't equal to Aij* + Bij*. A*+B* is a matrix. Aij* + Bij* is a number. It's not correct to say the two are equal. A clue is that neither i nor j appear on the lefthand side of the equation.

    It's probably been established (or assumed) in your class that if z and w are complex numbers, then (z+w)* = z* + w*. It's not "obvious" that the same relationship holds with a completely different type of object. When you write A*+B*, the + and * mean completely different things than the + and * in z* + w* because A and B are completely different mathematical objects than z and w. Recognizing this should help you understand what you need to show.
     
  15. Sep 19, 2015 #14

    RJLiberator

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    Good post Vela.

    What I got from this is that I need to show that the ij'th components from each side are equal.


    We actually proved that the conjugate of a sum of numbers is the sum of the conjugates earlier in the assignment, so I will use that to finish off this proof.
    I think you are right with where the confusion was, and I now feel very good about this proof.

    The right hand side is pretty straight forward when looking at the ij'th components.
    The left hand side, I have to first say that I am looking at the ij'th component, and then apply the earlier proof, to show that the left hand side is indeed equal to the right hand side.

    And this makes 100% sense to me now .
    Kind Regards
     
  16. Sep 19, 2015 #15

    vela

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    Glad it was helpful. After I posted, I reread your original post and wondered if perhaps you were clear on that and were just being a little sloppy, but I remember being confused the same way when I was first encountered proofs like these.
     
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