 Solution

A matrix $R$ belongs to the special orthogonal group in four dimensions precisely whenDifferentiating the orthogonality relation along a path $R(ϵ)=I+ϵX+O({ϵ}^2)$ gives $X^{\mathsf{T}}+X=0$. Thus its Lie algebra consists of real antisymmetric four-by-four matrices. Such a matrix has one free entry for each unordered coordinate pair, giving dimension $\left(\genfrac{}{}{0ex}{}{4}{2}\right)=6$.

Each displayed generator has a single two-coordinate antisymmetric block. The $K$ matrices cover planes $(1,2),(1,3),(1,4)$; the $L$ matrices cover $(3,4),(2,4),(2,3)$, with their printed orientations. These are all six distinct planes. Therefore the generators are linearly independent and span the full antisymmetric space: they form a basis of $\mathfrak{so}(4)$.

Conversely, exponentiating any real antisymmetric $X$ gives $(e^X{)}^{\mathsf{T}}{e}^X=I$ and $\det (e^X)=e^{\mathrm{tr}X}=1$. This verifies that the infinitesimal matrices generate proper rotations rather than reflections.