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Transpose of a matrix with mixed indices

  1. Jan 4, 2012 #1
    Hi!
    Given a matrix A of elements [itex]A_i\;^j[/itex], which is the right transpose:
    [itex]A_j\;^i[/itex]
    or
    [itex]A^j\;_i[/itex]
    ?
     
  2. jcsd
  3. Jan 4, 2012 #2

    Fredrik

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    Assuming that you mean that ##A_i{}^j## is what's on row i, column j, then the transpose of the matrix has ##A_j{}^i## on row i, column j.
     
  4. Jan 5, 2012 #3
    Uhm... I have this equation:
    [tex]\Lambda^T g \Lambda = g[/tex]
    in index notation:
    [tex]
    \left( \Lambda^T \right)^{\mu}\;_{\rho} g^{\rho}\;_{\alpha}\Lambda^{\alpha}\;_{\nu}=
    g^{\mu}\;_{\nu}
    [/tex]
    now,
    [tex]
    \left( \Lambda^T \right)^{\mu}\;_{\rho}=\Lambda^{\rho}\;_{\mu}
    [/tex]
    right?
    And so I get:
    [tex]
    \Lambda^{\rho}\;_{\mu} g^{\rho}\;_{\alpha}\Lambda^{\alpha}\;_{\nu}=
    g^{\mu}\;_{\nu}
    [/tex]
    Is this equation correct? (I don't think so, because the positions of the indices on the both sides are not correct)
     
  5. Jan 5, 2012 #4

    Fredrik

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    Actually, in this context (special relativity), it's conventional to write row ##\rho##, column ##\alpha## of ##g## as ##g_{\rho\alpha}##.

    The convention for ##g^{-1}## is that row ##\rho##, column ##\alpha## is written as ##g^{\rho\alpha}##.

    Also note that when you multiply the original equation by ##g^{-1}## from the left, you find that $$\Lambda^{-1}=g^{-1}\Lambda^Tg.$$ Row ##\rho##, column ##\alpha## of this matrix is written as
    $$(\Lambda^{-1})^\rho{}_\alpha =(g^{-1}\Lambda^Tg)^\rho{}_\alpha =g^{\rho\beta}\Lambda^\mu{}_\beta g_{\mu\alpha}=\Lambda_\alpha{}^\rho.$$
     
    Last edited: Jan 5, 2012
  6. Jan 5, 2012 #5
    here's my 2 cents.

    suppose i have two vector spaces (say X and Y) and a linear map f:X->Y
    then you automatically get a map f*:Y*->X*
    usually called the transpose or pullback. it's defined in the obvious way. let ω in Y*
    then (f*ω)(x) = ω(fx).

    if we have an inner product on X and Y we have an identification of X*
    with X and Y* with Y. we can use this to define the transpose map
    fT: Y ->X.

    Specifically, define θ:X->X* by (θx)(x') = (x,x') for all x' in X.
    similarly let ψ:Y->Y* {(ψy)(y')=(y,y')}. We put
    fT = θ-1f*ψ.

    or θfT = f*ψ.
    then for any y we have
    θfT(y) = f*ψ(y)
    both sides are elements of X* so that we can compare them
    by their action on an arbitrary x in X
    θfT(y)[x] = (fTy, x)
    f*ψ(y)[x] = ψ(y)( fx ) = (y, fx)

    so we have a coordinate independent definition of transpose for
    a map between two inner product spaces.
    Take a basis for X and Y.

    And find the components of any map f:X->Y by
    (fx)i = fij xj

    we have
    (y, fx) = gijyi(fx)j = gijyifjkxk

    and(fTy, x) = Gij(fTy)ixj= Gij(fT)ikykxj.

    Reindex the dummy variables and compare:
    gijfjk = Gjk(fT)ji.

    or using the convention that the inverse of G has components Gij

    (fT)ij = gjkfklGli.

    using the convention that g (or G) raises or lowers indices we have

    (fT)ij = fji.

    -----------------------

    This is just a longwinded way to say
    gijfjk = Gjk(fT)ji is the same as fik = (fT)ki,
    then make indices match on both sides of the equation.
     
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