Why have we done all this? Well, first of all, note that, due to antisymmetry, $\bT$ has precisely 6 independent components. Next, compute $\bT'$, using matrix multiplication and the fundamental hyperbolic trig (4) of §4.1. As you should check for yourself, the result is \begin{equation} T'{}^{\mu\nu} = \pmatrix{ 0& a'& b'& c'\cr -a'& 0& f'& -e'\cr -b'& -f'& 0& d'\cr -c'& e'& -d'& 0\cr } \end{equation} where \begin{eqnarray} a' &=& a \\ b' &=& \cish b-f \\ c' &=& \cish c+e \\ d' &=& d \\ e' &=& \cish e+c \\ f' &=& \cish f-b \end{eqnarray} The first three of these are similar to the transformation rule for the electric field, and the remaining three are similar to the transformation rule for the magnetic field!

We conclude that the electromagnetic field is described by an antisymmetric, rank 2 tensor of the form \begin{equation} F^{uv} = \pmatrix{ 0& \cm\,{E^x}& \cm\,{E^y}& \cm\,{E^z}\cr -\cm\,{E^x}& 0& B^z& -B^y\cr -\cm\,{E^y}& -B^z& 0& B^x\cr -\cm\,{E^z}& B^y& -B^x& 0\cr } \end{equation} which is known as the electromagnetic field tensor.