Roughly speaking, tensors are like vectors, but with more components, and hence more indices. We will only consider one particular case, namely rank 2 contravariant tensors, which have two “upstairs” indices. Such a tensor has components in a particular reference frame which make up a $4\times4$ matrix, \begin{equation} T^{\mu\nu} = \pmatrix{ T^{00}& T^{01}& T^{02}& T^{03}\cr T^{10}& T^{11}& T^{12}& T^{13}\cr T^{20}& T^{21}& T^{22}& T^{23}\cr T^{30}& T^{31}& T^{32}& T^{33}\cr } \end{equation} How does the tensor $\bT$ transform under Lorentz transformations? Well, it has two indices, each of which must be transformed. This leads to a transformation of the form \begin{equation} T'{}^{\mu\nu} = \Lambda^\mu{}_\rho \Lambda^\nu{}_\sigma T^{\rho\sigma} = \Lambda^\mu{}_\rho T^{\rho\sigma} \Lambda^\nu{}_\sigma \end{equation} where the second form (and the summation convention!)\ leads naturally to the matrix equation \begin{equation} \bT' = \bL \bT \bL^t \end{equation} where $t$ denotes matrix transpose.

Further simplification occurs in the special case where $\bT$ is antisymmetric, that is \begin{equation} T^{\nu\mu} = -T^{\mu\nu} \end{equation} so that the components of $\bT$ take the form \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}