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Algebra: unitary mappings

  1. Nov 26, 2006 #1
    i need to show that the unitary maps form a group under multiplication. ive never had algebra (im in linear algebra) so this is the first time ive seen this idea of 'groups'. i tried to look stuff up, so i think i see now that i need to show: 1) identity is an element 2) if x,y are elements of the group then the product xy is in the group 3) if x is an element, then its inverse is in the group.

    I'm not really sure how to show these things since I'm new to this. :bugeye: All I really have to work with is that a unitary mapping M satisfies ll Mx - My ll = ll x - y ll for all x,y.

    Can you help me get started? Thanks, Sarah :)
  2. jcsd
  3. Nov 26, 2006 #2


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    Technically, you also need to check that multiplication is associative, but that can probably be taken for granted here. So, show:

    1) The identity map is unitary [this is really, really easy]
    2) If X and Y are unitary, then XY is unitary [the only thing easier than this is 1)]
    3) If X is unitary, then a) X-1 exists and b) X-1 is unitary.

    You must be able to do 1) and 2) without help, and at least make some effort and show some work one 3a) and b) before anyone should help you here.
  4. Nov 28, 2006 #3
    I think I'm getting there now.

    Let U be the collection of unitary maps. I need to show that U is a group under multiplication.

    1. the identity is an element of U because:
    I*I = I , therefore I is an element of U.

    2. Let M1, M2 be elements of U. I need to show that M1M2 is an elt. of U:

    (M1M2)*(M1M2) = I
    Now the LHS is = (M2*M1*)(M1M2)
    = M2*(M1*M1)M2
    = M2* (I) M2
    = M2*M2
    = I

    therefore M1M2 is an elt. of U

    3. for the inverse, I can just refer to another hw problem i was assigned before this where we showed this. I got that one.
  5. Nov 28, 2006 #4
    I'm obviously using here that a unitary map M satisfies M*M = I. I didn't realize I could use that for this problem. Then I did and it become much clearer.
  6. Nov 28, 2006 #5


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    It's just as easy doing it using the definition that M is unitary iff ||Mx - My|| = ||x - y|| for all x and y.

    1. ||Ix - Iy|| = ||x - y|| so I is unitary
    2. If M, N are unitary ||MNx - MNy|| = ||Nx - Ny|| = ||x - y||, so MN is unitary
    3. If M is unitary and M-1 exists, then:

    ||M-1x - M-1y||
    = ||MM-1x - MM-1y|| (since M is unitary)
    = ||Ix - Iy||
    = ||x - y||

    so M-1 is unitary. To prove M-1 exists, suppose it didn't. Then there would exist non-zero x such that Mx = 0 (because M is linear, which you forgot to mention in your opening post). But then:

    = ||0 - 0||
    = ||Mx - M0|| (since 0 = M0 and 0 = Mx)
    = ||x - 0|| (since M is unitary)
    > 0 (since x is non-zero)

    0 > 0 is a contradiction, so M does have an inverse.
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