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The Product of two Unitary Matrices is Unitary Proof

  1. Sep 20, 2015 #1

    RJLiberator

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    Gold Member

    1. The problem statement, all variables and given/known data
    Show that the product of two nxn unitary matrices is unitary. Is the same true of the sum of two nxn unitary matrices?

    2. Relevant equations
    Unitary if A†A=I
    Where † = hermitian conjugate
    I = identity matrix.

    3. The attempt at a solution

    We have the condition: (AB)†(AB)=I
    I can then apply summation notation for the elements of the matrices

    [itex](AB)^†(AB) = \Big( \sum_{k=1}^j(AB)^†_{ij}(AB)_{ij} \Big)[/itex]


    Now, the idea, I suppose, is to manipulate the sum so that we see A^† *A and B^†*B and we can conclude that since A and B are unitary, then A*B is unitary.

    This seems coherent, and beautiful.

    I assume summation notation is needed to make this distinction. Correct?

    And if this is so, I will undo the hermitian conjugate first and then manuever the pieces of the elements.
    Correct?
     
  2. jcsd
  3. Sep 20, 2015 #2

    RJLiberator

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    Gold Member

    Let me write out my proof for checking:

    1. [itex](AB)^†(AB)_{ij} = \Big( \sum_{k=1}^j(AB)^†_{ij}(AB)_{ij} \Big)[/itex]
    by element notation
    2. [itex](AB)^†(AB)_{ij} = \Big( \sum_{k=1}^j(AB)^*_{ji}(AB)_{ij} \Big)[/itex]
    by hermitian conjugate definition
    3. [itex](AB)^†(AB)_{ij} = \Big( \sum_{k=1}^jA^*_{ji}B^*_{ji}A_{ij}B_{ij} \Big)[/itex]
    By proven earlier property that (AB)*=A*B* and earlier proposition that ABij =AijBij
    4.[itex](AB)^†(AB)_{ij} = \Big( \sum_{k=1}^jA^*_{ji}A_{ij}B^*_{ji}B_{ij} \Big)[/itex]
    By operations, since we are dealing with components
    5.[itex](AB)^†(AB)_{ij} = \Big( \sum_{k=1}^jI_{ij}I_{ij} \Big)[/itex]
    By earlier conditions stating these are two unitary matrices
    6. Therefore I_ij = I_ij

    And walouh! First part, done.

    Any errors in my thinking?
     
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