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Homework Help: Det(tA) = Det(A)

  1. Dec 9, 2009 #1
    1. The problem statement, all variables and given/known data
    Prove that the determinant of a unitary matrix is +/-1



    2. Relevant equations
    <Av,Aw>=<v,w>
    Det(AB)=Det(A)*Det(B)



    3. The attempt at a solution
    Alright, I am aware that <Av,Aw>=<v,w> => A(tA)=I and (tA)A=I so Det(A(tA))=Det(I)=1 thus
    Det(A)*Det(tA)=1. However, this is where I am stuck, I am aware that in some cases Det(tA)=Det(A), however this is not obvious in this case, as I cannot take tA to be the elementary defintion of switching rows and columns. How can I prove that Det(tA)=Det(A) ONLY from the fact that <Aw,v>=<w,(tA)v>. This inner product can be absolutely any symmetric bilinear form, so do not assume I mean the dot product.
     
    Last edited: Dec 9, 2009
  2. jcsd
  3. Dec 9, 2009 #2
    Re: Det(tA)=Det(A)

    I guess it should be |det(A)|=1, rather than det(A)=+/-1.
     
  4. Dec 9, 2009 #3

    Dick

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    Re: Det(tA)=Det(A)

    If you are trying to prove that the determinant of a unitary matrix is +/-1 then you MUST mean REAL unitary. It's not true for complex unitary. That means your tA is really just a plain old transpose.
     
  5. Dec 9, 2009 #4
    Re: Det(tA)=Det(A)

    Even for some random inner product, I think this is what I don't understand, why must it be the normal transpose for any completely random product, it only seems true for R^n with the standard dot product. Like for example, taking the space to be the space to be Pn(X) and <,> to be the integral of f*g.
     
  6. Dec 9, 2009 #5
    Re: Det(tA)=Det(A)

    Thank you for your responses, I think I have proved that for real unitary matrices, the transpose really is just the transpose. here is my proof if anyone can check or is interested.

    Lemma: If U is unitary real, then it commutes with the matrix rep. of a nondegenerate bilinear form.
    Pf: Let U be unitary, and let B be a nondegenerate bilinear form. Then

    B(Uw,UV)=t(Uw)B'(Uv) for some matrix representation of B

    so (tw*tU)B'(Uv)=(tw)*(tU*B'*U)*(v)

    on the other hand B(Uw,Uv)=B(w,v), thus (tU*B'*U)=B',

    however, since U is unitary, tU=U^-1, so (U^-1)*B'*U=B' and thus multiplying both sides by U, B'*U=U*B'. Thus U commutes with B'.

    So If B is a symmetric bilinear form, tU really does equal the transpose.

    Pf: Consider B(Uw,v)=B(v,Uw)=(tv)B'(Uw)=(tv)(B'U)w=(tv)(UB')w=t(tU*v)B'w=B((tU)v,w).

    Sorry about the horrible notation due to my not using latex. Again, thank you Dick for pointing me towards this proof.
     
  7. Dec 9, 2009 #6

    Dick

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    Re: Det(tA)=Det(A)

    Ok, you're welcome! Glad you convinced yourself. I would conceptualize it by picking an orthonormal basis for the form using Gram-Schmidt, so <ei,ej>=delta_ij. So the matrix of U is U_ij=<ei,Uej>. So tU_ij=<ei,tUej>=<Uei,ej>=<ej,Uei>=U_ji. Or did I get my indices backwards? Doesn't matter, you get my point, right? So in the complex case it's pretty obvious the adjoint is the conjugate transpose.
     
  8. Dec 9, 2009 #7
    Re: Det(tA)=Det(A)

    Yes, I think this revealed a lot to me. I think that I was just being very cautious, because the notation was very suggestive, and so I went into some sort of lockdown mode. I do like the way of conceptualizing this that you presented, and now I do understand that in the complex case it is conjugate transpose, so thank you much.
     
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