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Hermitian Identity

  1. Oct 11, 2011 #1
    1. The problem statement, all variables and given/known data

    Show that (AB)^+ = A^+ B^+ using index notation


    2. Relevant equations

    + is the Hermitian transpose


    3. The attempt at a solution

    I know that AB = Ʃa_ik b_kj summed over k

    so (AB)^+ = (Ʃa_ik b_kj)^+ = Ʃ (a_ik b_kj)^+ = Ʃ (a_ik)^+(b_kj)^+ = A^+ B^+

    I am not really sure if this makes sense, I don't know if it is acceptable to distribute the transpose within the sum.
     
  2. jcsd
  3. Oct 11, 2011 #2

    Dick

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    (AB)^+ is equal to (B^+)(A^+), not (A^+)(B^+). That may be a sign something is going wrong. (A^+)_ij=(A_ji)*, where * is complex conjugate. Start from there.
     
  4. Oct 12, 2011 #3
    That helped.

    So now using the definition (usually a good thing) I have:

    (AB)^+_ij = [AB_ji]^* = A_ji ^* B_ji^* = B_ij^+ A_ij^+ = B^+ A^+

    I know that the complex conjugate is distributive can I just assume that for the proof?
     
  5. Oct 12, 2011 #4

    Dick

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    Yes, you can use (xy)^*=(x^*)(y^*). That's fine. But now you've lost the matrix product part. AB is a product. (AB)_ij isn't equal to A_ij B_ij.
     
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