I have heard relativistics only very compressed during my student time. Now I looked up the definitions again and a question comes into my mind:
A contravariant vector is transformed like this: $(a^\mu)'=L_{\mu\lambda}a^\lambda$ wherein $L_{\mu\lambda}$ is the Lorentz matrix. A covariant vector is transformed like this: $(a_\mu)'=L^{-1}_{\lambda\mu}a_\lambda=L^{T}_{\lambda\mu}a_\lambda=L_{\mu\lambda}a_\lambda$. Does this mean no matter if I have a co- or contravariant four vector I use the same matrix to transform them from one coordinate system to the other?
And one general question: Why do I need co- AND contravariant four vectors? Isn't one enough?
EDIT: My definitions
I think I found the solution. We do not have the common definitions of the four vectors! I have a definition like this:
$a^\mu=(a_1,a_2,a_3,\mathrm{i}ct)$
The normal definition is:
$a^\mu=(a_1,a_2,a_3,ct)$
Then, regarding the answers, it doesn't matter where I put the indices of the Lorentz matrix, does it? Within the new definition, the Lorentz matrix becomes non-symmetric
$L=\left(\begin{array}{ccc}\cosh \theta &0&0&\mathrm{i}\sinh \theta \\ 0&1&0&0\\0&0&1&0\\ -\mathrm{i}\sinh&0&0&\cosh \theta\end{array}\right)$
In contrast, the normal Lorentz matrix is
$L=\left(\begin{array}{ccc}\cosh \theta &0&0&\sinh \theta \\ 0&1&0&0\\0&0&1&0\\ \sinh&0&0&\cosh \theta\end{array}\right)$
Now we see that indeed $(a_\mu)'=L^{-1}_{\lambda\mu}a_\lambda=L^{T}_{\lambda\mu}a_\lambda=L_{\mu\lambda}a_\lambda$. So in this representation, the transformation of co- and contravariant four vectors are done by the same matrix. Can anyone confirm?