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Def. of unitary

  1. Apr 12, 2004 #1

    turin

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    I have read two definitions of unitary.
    A is unitary if:
    #1: det(A) = 1
    or
    #2: AadjointA = I

    Are these definitions equivalent?
     
  2. jcsd
  3. Apr 12, 2004 #2

    matt grime

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    Absolutely not. And you want to be careful with adjoint, better to specify that you mean A* the conjugate trasnpose.

    1 is sufficient for A to be SL

    2 says, with adjoint understood as above it is unitary.

    there are of course non-unitary matrices of det 1. exercise find a simple upper triangular real (integer) matrix to show this.
     
  4. Apr 12, 2004 #3

    turin

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    What is the difference between "adjoint" and "conjugate transpose?"




    What is "SL?"




    What is an "upper triangular" matrix? If it is what I think it is, then the determinant is just the product of the trace elements. This will not be a different value for the transpose of the matrix.
     
    Last edited: Apr 12, 2004
  5. Apr 12, 2004 #4

    matt grime

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    Adjoint is context dependent. if the matrices are real then adjoint is just transpose - part of the definition for unitary is that it is complex, and preserves an hermitian form (which is often how you'll see it defined). Just writing adjoint without specifying we mean complex at some point is omitting some information. One could argue that A^{adjoint}A =I defines the orthogonal (real) matrices.

    SL(F) is the special linear group - the group of matrices with determinant 1 (with entries in the field F)

    upper triangular means all entries below the main diagonal are zero.

    You can do the exerices with a diagonal matrix too, though obviously not one with integer coefficients.
     
  6. Apr 12, 2004 #5

    turin

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    matt grime,
    I did not understand the point you were trying to make about "adjoint." If the matrices are real, then the adjoint and transpose are the same thing, I get that. I don't see how this makes "adjoint" and "transpose conjugate" different (since the conjugate of a real number is itself).
     
    Last edited: Apr 12, 2004
  7. Apr 12, 2004 #6

    matt grime

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    If you just say that the matrix A satisfies A^{adjoint}A=I, we do not know if you mean orthogonal or unitary, at least indicate that you mean complex matrices.
     
  8. Apr 12, 2004 #7

    turin

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    This just seems a little strange to me. If the components of A are completely real-valued, then even though the condition were to specify that A is orthogonal, it would also be unitary. If the components were not all completely real-valued, are you saying that "adjoint" can strictly mean "transpose" in this case without complex conjugation?

    Please don't think I'm trying to be difficult. I was just taught that "adjoint" always means "transpose" and "complex conjugate" and I would like to be aware of any variation in usage for the sake of future conversations with the math people.
     
    Last edited: Apr 12, 2004
  9. Apr 13, 2004 #8

    matt grime

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    That is waht it means if the matirx is complex. Don't take this the wrong way, but as you asked if unitary meant det is 1 I assume you've not done much maths, thus I want to see that you understand that unitarity is a property of compelx matrices, and just saying adjoint doesn't do that.

    Let C be some space (what kind we won't say), with a bilinear map (?,?) on the elements of that space. THe (right) adjoint of a map T:C-->C is a map T* satisfying (TX,Y)=(X,T*Y)

    it occurs in many places, in the case of a complex vector space it means transpose conjugate and the bilinear map is the natural hermitian inner product.
     
  10. Apr 13, 2004 #9

    turin

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    No problem. And, you are correct.




    I don't follow this. What is a "bilinear map?" I'm assuming that X and Y are like vectors and that T is like an operator?

    I don't understand the following notation

    T:C-->C
     
  11. Apr 13, 2004 #10

    matt grime

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    I was trying not to have to specify what any of the things is, but C is some "space" X and Y are some elements of the space (it needn't be a vector space with vectors just a couple of sets with elements) and, T is a map from X to Y - that is what T:C--->C means.

    (?,?) is a just some way of paring up elements of C to get something in some other set, such as the real or complex numbers. (I didn't mean to say bilinear, just binary ie two inputs).

    *The* model for this is, yes, vector spaces (over R or C, this C isn't the other C in the post unfortunately) and vectors and linear maps and the inner product (or hermitian product), but there are other places where adjoint gets used.

    The main thing is if you ask:

    What is the group of matrices described by the law A^{adjoint}A=I=AA^{adjoint}?

    Then the answer is either the unitary group OR the orthogonal group depending on whether the matrices are real or complex. One is a subgroup of the other. It is nitpicking but if you don't make precise the terms you use then maths becomes hard.
     
    Last edited: Apr 13, 2004
  12. Apr 13, 2004 #11
    Or if the matrices are quaternionic, then we have the symplectic group
     
  13. Apr 13, 2004 #12
    You might be thinking of the group SU(n). the "S" in SU(n) means special linear, which means det(A) = 1. the "U" in SU(n) means unitary, which means A^adjoint*A=1 (where here, "adjoint" mean transpose conjugate)

    so a matrix in SU(n) satisfies both of your conditions. a matrix that satisfies only condition #1 (but not necessarily condition #2) is in SL(n,C), SL means special linear.

    if it satisfies condition #2 (but not necessarily condition #1) then the matrix is in U(n), for unitary.

    then the intersection of SL(n,C) and U(n) is SU(n). matrices that satisfy both condition #1 and #2
     
  14. Apr 13, 2004 #13

    matt grime

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    That's just evil, but as the quaternions aren't a field I don't feel too bad for not addressing them.
     
  15. Apr 13, 2004 #14

    turin

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    Sorry. I'm probably the wrong guy for you to be helping, then.




    I think that it makes sense to me in this way. But I am still confused by T:C-->C. That is, it seems like T should be the identity in this case.




    I definitely don't mind the nitpicking. Thanks for the help.
     
  16. Apr 13, 2004 #15

    turin

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    lethe,
    What does the C mean in SL(n,C)?
     
  17. Apr 13, 2004 #16

    Janitor

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    Here is something that Zee says in his book on Q.F.T.

     
  18. Apr 13, 2004 #17

    matt grime

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    Lethe's C is the complex numbers. My C was supposed to suggest Category.

    No one will ever agree on notation it seems. In Hilbert space theory adjoint is often just written * because overline denotes complex conjugation (damn physicists mucking it up), and who on earth wants to just take the conjugate of a matrix?
     
    Last edited: Apr 13, 2004
  19. Apr 15, 2004 #18
    how about sesquilinear? i like that word.
     
  20. Apr 15, 2004 #19
    you're too discriminating! you'll never cut it as a physicist!

    anyway, in case anyone is interested see this page for an explanation of the following analogy:

     
  21. Apr 15, 2004 #20
    my favorite notation convention (one that is, i think, shared by many physicists) is for [itex]z[/itex] and [itex]\overline{z}[/itex] to be your independent complex variables, and restrict yourself to the 2d plane where [itex]z^*=\overline{z}[/itex]
     
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