Transpose of a matrix with mixed indices

  • Thread starter eoghan
  • Start date
  • #1
200
1

Main Question or Discussion Point

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]
?
 

Answers and Replies

  • #2
Fredrik
Staff Emeritus
Science Advisor
Gold Member
10,851
408
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.
 
  • #3
200
1
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)
 
  • #4
Fredrik
Staff Emeritus
Science Advisor
Gold Member
10,851
408
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:
  • #5
185
3
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.
 

Related Threads on Transpose of a matrix with mixed indices

  • Last Post
Replies
1
Views
9K
  • Last Post
Replies
1
Views
9K
  • Last Post
Replies
1
Views
4K
  • Last Post
Replies
2
Views
5K
  • Last Post
Replies
3
Views
336
Replies
5
Views
1K
  • Last Post
Replies
2
Views
6K
Replies
6
Views
3K
Replies
6
Views
1K
  • Last Post
Replies
1
Views
1K
Top