Lorentz-transformation I don't understand how to derive the matrix representing the Lorentz-transformation given two systems S and S':
$$x' = \Lambda x$$
these transformations do not leave the differences $\Delta x^\mu$ unchanged, but multiply them also by the matrix $\Lambda$:
$$(\Delta s)^2 = (\Delta x)^T \eta (\Delta x) = (\Delta x')^T \eta (\Delta x') = (\Delta x)^T \Lambda^T \eta \Lambda (\Delta x) \tag{1.26}$$
and therefore
$$\eta = \Lambda^T \eta \Lambda$$
I don't understand the mathematical passages in eq. 1.26 particularly:


*

*why is it needed to multiply by $\Lambda$?

*The role of the transpose symbol


I know that the space-time interval is given by 
$$(\Delta s)^2 = \eta_{\mu \nu}\Delta x^\mu\Delta x^\nu$$
and I understand that the metric given by $\eta$ should be the same in every reference system.
 A: To answer your question:


*

*If you look at $\mathrm{(1.26)}$ carefully, you will see that we have
$$(\Delta s)^2 = (\Delta x )^T \eta(\Delta x) = (\Delta x')^T \eta(\Delta x')$$
This is equating the line element in the two different inertial frames. Notice the expression on the right hand side is a function of $\Delta x'$ not $\Delta x$. Using $\Delta x' =  \Lambda \Delta x$ and $(\Delta x')^T = (\Delta x)^T \Lambda^T$ you can see where the both $\Lambda$ and $\Lambda^T$ have come from.

*The transpose symbol is because the line element must be a scalar, so the "matrix" that represents both $\Delta x$ terms must have inverse dimensions to the "matrix" that represents the metric. I personally don't like this formalism, and if you go on to do general relativity you will be introduced to tensors and the Einstein summation convention. In this formalism it is a lot clearer because the line element is given as
$$ds^2=g_{\mu \nu} dx^\mu dx^\nu$$
where $g_{\mu \nu}$ is the metric. In this notation you can clearly see that the line element is a scalar and that metric is a (0,2) tensor so $dx^\mu dx^\nu$, if treated like a single object like it is in your question, must be a rank (2, 0) tensor.
I must admit I personally do not like the form given in $\mathrm{(1.26)}$ since I feel it is confusing as to what $\Delta x$ is and why, as you asked, we want to take the transpose. So I will write the same derivation in the notation used in general relativity, hopefully you may find this clearer.
We have the line element in the two frames as
$$ds^2 =  \eta_{\mu \nu}dx^\mu dx^\nu = \eta'_{\mu' \nu'} dx^{\mu'} dx^{\nu'}$$ 
where we are working in Minkowski space (special relativity) so  $\eta = \eta'$. The $\mu'$ and $\nu'$ are to denote that these are new indices in the new frame. These are given by the Lorentz transformation
$$dx^{\mu'} = \Lambda^{\mu'}{}_\mu dx^\mu.$$ Notice that here the transmormation "matrix" is actually a (1,1) tensor. Einstein summation convention tells us we sum over the repeated $\mu$ so what the only free index on the right-hand side is $\mu'$, just like on the left-hand side. Substituting this into the above gives
$$\eta_{\mu \nu}dx^\mu dx^\nu = \eta'_{\mu' \nu'} \Lambda^{\mu'}{}_\mu \Lambda^{\nu '}{}_\nu dx^\mu dx^\nu$$
So,
$$\eta_{\mu \nu} = \Lambda^{\mu'}{}_\mu \Lambda^{\nu'}{}_{\nu} \eta'_{\mu' \nu'}$$
Which is the same as
$$\eta'=\Lambda^T\eta\Lambda.$$
A: 
The role of the transpose symbol

The vectors are usually represented by the column matrix. $$\Delta x= \begin{bmatrix}c\Delta t \\ \Delta x \\\Delta y\\ \Delta z \end{bmatrix} .$$ Thus $$\Delta x^T= \begin{bmatrix}c\Delta t & \Delta x &\Delta y& \Delta z \end{bmatrix} .$$ The invariant interval in matrix representation is given by $$\Delta s^2=\Delta x^T\eta\Delta x=\begin{bmatrix}c\Delta t & \Delta x &\Delta y& \Delta z \end{bmatrix}\begin{bmatrix}
1& 0 & 0 & 0\\
0 & -1 & 0 & 0 \\
0 & 0 & -1 & 0 \\
0& 0 & 0 & -1
\end{bmatrix}\begin{bmatrix}c\Delta t \\ \Delta x \\\Delta y\\ \Delta z \end{bmatrix}$$
$$=(c\Delta t)^2 - \Delta x^2 -\Delta y^2 - \Delta z^2$$

Why is it needed to multiply by Λ?

Because the Lorentz transformation equation is given by $$\Delta x' = \Lambda \Delta x$$
substituting this in the interval $(\Delta s)^2 = (\Delta x')^T \eta (\Delta x')$  we get
$$ (\Delta s)^2= (\Lambda\Delta x)^T \eta (\Lambda\Delta x)=\Delta x^T\Lambda^T\eta \Lambda\Delta x$$
This is same as  $(\Delta x)^T \eta (\Delta x)$ . Analyse it with the above equation, you will obtain $\eta = \Lambda^T \eta \Lambda$
