# Transpose of a matrix with mixed indices

1. Jan 4, 2012

### eoghan

Hi!
Given a matrix A of elements $A_i\;^j$, which is the right transpose:
$A_j\;^i$
or
$A^j\;_i$
?

2. Jan 4, 2012

### Fredrik

Staff Emeritus
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. Jan 5, 2012

### eoghan

Uhm... I have this equation:
$$\Lambda^T g \Lambda = g$$
in index notation:
$$\left( \Lambda^T \right)^{\mu}\;_{\rho} g^{\rho}\;_{\alpha}\Lambda^{\alpha}\;_{\nu}= g^{\mu}\;_{\nu}$$
now,
$$\left( \Lambda^T \right)^{\mu}\;_{\rho}=\Lambda^{\rho}\;_{\mu}$$
right?
And so I get:
$$\Lambda^{\rho}\;_{\mu} g^{\rho}\;_{\alpha}\Lambda^{\alpha}\;_{\nu}= g^{\mu}\;_{\nu}$$
Is this equation correct? (I don't think so, because the positions of the indices on the both sides are not correct)

4. Jan 5, 2012

### Fredrik

Staff Emeritus
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
5. Jan 5, 2012

### qbert

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.