in which the elements fulfill the relation $f_{kh}=-f_{hk}$, is called an *alternating* matrix. These relations say that the transposed matrix ${\bar {f}}=-f$. Then by $f^{*}$ will be the *dual*, alternating matrix

(35) |
$f^{*}=\left|{\begin{array}{cccc}0,&f_{34},&f_{42},&f_{23}\\f_{43},&0,&f_{14},&f_{31}\\f_{24},&f_{41},&0,&f_{12}\\f_{32},&f_{13},&f_{21},&0\end{array}}\right|$, |

Then

(36) |
$f^{*}f=f_{32}f_{14}+f_{13}f_{24}+f_{21}f_{34}$, |

*i.e.* We shall have a 4✕4 series matrix in which all the elements except those on the diagonal from left up to right down are zero, and the elements in this diagonal agree with each other, and are each equal to the above mentioned combination in (36).

The determinant of *f* is therefore the square of the combination, by $Det^{\frac {1}{2}}f$ we shall denote the expression

(37) |
$Det^{\frac {1}{2}}f=f_{32}f_{14}+f_{13}f_{24}+f_{21}f_{34}$ |

4°. A linear transformation

(38) |
$x_{h}=\alpha _{h1}x'_{1}+\alpha _{h2}x'_{2}+\alpha _{h3}x'_{3}+\alpha _{h4}x'_{4}\qquad (h=1,2,3,4)$ |

which is accomplished by the matrix

${\mathsf {A}}=\left|{\begin{array}{cccc}\alpha _{11},&\alpha _{12},&\alpha _{13},&\alpha _{14}\\\alpha _{21},&\alpha _{22},&\alpha _{23},&\alpha _{24}\\\alpha _{31},&\alpha _{32},&\alpha _{33},&\alpha _{34}\\\alpha _{41},&\alpha _{42},&\alpha _{43},&\alpha _{44}\end{array}}\right|$,
will be denoted as the transformation ${\mathsf {A}}$

By the transformation ${\mathsf {A}}$, the expression

$x_{1}^{2}+x_{2}^{2}+x_{3}^{2}+x_{4}^{2}$
is changed into the quadratic form

$\Sigma a_{hk}x'_{h}x'_{k}\qquad (h,k=1,2,3,4)$