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Traceless hermitian matrices form groups?

  1. Nov 12, 2008 #1
    1. is the set of nxn traceless hermitian matrices under addition a group?
    2. is the set of nxn traceless hermitian matrices under multiplication a group?
    3. is the set of nxn traceless non-hermitian matrices under addition a group?

    question 1-I thought that traceless means trace=0 is this right? so what would the identity element be? it can't be the null matrix because it doesnt have an inverse, can anyone help? I haven't got around to the other questions but help is probably needed coz i dont like matrices
  2. jcsd
  3. Nov 12, 2008 #2
    I just realised in the first quesiton, the composition law is actually addition, so that makes the inverse of the identiy just putting a minus sign on all of its elements, which doesnt change the diagonal, which mean its still traceless, so it must be a group.

    for the second question closure isn't satisfied , the third one im not sure what to do...
  4. Nov 12, 2008 #3
    non-hermitian matrices don't include the identity.
  5. Nov 12, 2008 #4
    yes of course, thanks a lot
  6. Nov 12, 2008 #5


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    Well, first of all, the identity element for addition is the matrix of all zeroes, not the identity matrix. Of course, this is also hermitian. But "non-hermitian" is often supposed to mean "not necessarily hermitian" rather than "definitely not hermitian". The answer depends on which meaning is implied.
  7. Nov 12, 2008 #6
    That was what I meant.

    "Not necessarily hermitian" just means all matrices. Then, there is no point in using such term.

    To me it seems safe to consider "non-hermitian" as "definitely not hermitian".
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