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Proof that U(n) is a group

  1. Nov 6, 2015 #1
    1. The problem statement, all variables and given/known data

    Show that the set of all ##n \times n## unitary matrices forms a group.

    2. Relevant equations

    3. The attempt at a solution

    For two unitary matrices ##U_{1}## and ##U_{2}##, ##x'^{2} = x'^{\dagger}x' = (U_{1}U_{2}x)^{\dagger}(U_{1}U_{2}x) = x^{\dagger}U_{2}^{\dagger}U_{1}^{\dagger}U_{1}U_{2}x = x^{\dagger}U_{2}^{\dagger}U_{2}x = x^{\dagger}x = x^{2}.##

    So, closure is obeyed.

    Matrix multiplication is associative.

    The identity element is the identity matrix.

    ##x'^{2} = (U^{-1}x)^{\dagger}(U^{-1}x) = x^{\dagger}(U^{-1})^{\dagger}U^{-1}x = x^{\dagger}(U^{\dagger})^{-1}U^{-1}x = x^{\dagger}(UU^{\dagger})^{-1}x = x^{\dagger}x = x^{2}##.

    So, the inverse of any unitary matrix is a unitary matrix.

    Is my answer correct?
    Last edited: Nov 6, 2015
  2. jcsd
  3. Nov 6, 2015 #2


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    2017 Award

    Staff: Mentor

    yep, and if you mention that the identity is clearly unitary and the inverse exists at all, e.g. because |det(U)| = 1 it'll be perfect
  4. Nov 6, 2015 #3
    Thanks! Got it!
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