= Solution
A <matrix> $R$ belongs to the <special orthogonal group> in four dimensions precisely when
$$
\boxed{R\in M_4(\mathbb R),\qquad R^{\mathsf T}R=I,\qquad\det R=1.}
$$
Differentiating the <orthogonality> relation along a path $R(\epsilon)=I+\epsilon X+O(\epsilon^2)$ 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 $\binom42=6$.
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: \b[they form a basis of $\mathfrak{so}(4)$].
Conversely, exponentiating any real antisymmetric $X$ gives $(e^X)^{\mathsf T}e^X=I$ and $\det(e^X)=e^{\operatorname{tr}X}=1$. This verifies that the infinitesimal <matrices> generate proper rotations rather than reflections.
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