For this matrix I shall use the shortened from lor.
Then if S is, as in (62), a spacetime matrix of the II. kind, by lor S' will be understood the 1✕4 series matrix
$\leftK_{1},\ K_{2},\ K_{3},\ K_{4}\right$
where
(64) 
$K_{k}={\frac {\partial S_{1k}}{\partial x_{1}}}+{\frac {\partial S_{2k}}{\partial x_{2}}}+{\frac {\partial S_{3k}}{\partial x_{3}}}+{\frac {\partial S_{4k}}{\partial x_{4}}}\qquad (k=1,2,3,4)$ 
When by a Lorentz transformation ${\mathsf {A}}$, a new reference system $x'_{1},\ x'_{2},\ x'_{3},\ x'_{4}$ is introduced, we can use the operator
$lor'=\left{\frac {\partial }{\partial x'_{1}}},\ {\frac {\partial }{\partial x'_{2}}},\ {\frac {\partial }{\partial x'_{3}}},\ {\frac {\partial }{x'_{4}}}\right$
Then S is transformed to $S'={\bar {\mathsf {A}}}S{\mathsf {A}}=\leftS'_{hk}\right$, so by lor' Sis meant the 1✕4 series matrix, whose element are
$K'_{k}={\frac {\partial S'_{1k}}{\partial x'_{1}}}+{\frac {\partial S'_{2k}}{\partial x'_{2}}}+{\frac {\partial S'_{3k}}{\partial x'_{3}}}+{\frac {\partial S'_{4k}}{\partial x'_{4}}}\qquad (k=1,2,3,4)$
Now for the differentiation of any function of (x y z t) we have the rule

${\frac {\partial }{\partial x'_{k}}}={\frac {\partial }{\partial x_{1}}}{\frac {\partial x_{1}}{\partial x'_{k}}}+{\frac {\partial }{\partial x_{2}}}{\frac {\partial x_{2}}{\partial x'_{k}}}+{\frac {\partial }{\partial x_{3}}}{\frac {\partial x_{3}}{\partial x'_{k}}}+{\frac {\partial }{\partial x_{4}}}{\frac {\partial x_{4}}{\partial x'_{k}}}$
$={\frac {\partial }{\partial x_{1}}}\alpha _{1k}+{\frac {\partial }{\partial x_{2}}}\alpha _{2k}+{\frac {\partial }{\partial x_{3}}}\alpha _{3k}+{\frac {\partial }{\partial x_{4}}}\alpha _{4k}$,

so that, we have symbolically
$lor'=lor\ ({\mathsf {A}}$
Therefore it follows that
(65) 
$lor'\ S'=lor({\mathsf {A}}({\mathsf {A}}^{1}S{\mathsf {A}}))=(lor\ S){\mathsf {A}}$, 
i.e., lor S behaves like a spacetime vector of the first kind.
If L is a multiple of the unit matrix, then by lor L will be denoted the matrix with the elements