Page:Scientific Papers of Josiah Willard Gibbs - Volume 2.djvu/72

118. Since all the antecedents or all the consequents in any dyadic may be expressed in parts of any three non-complanar vectors, and since the sum of any number of dyads having the same antecedent or the same consequent may be expressed by a single dyad, it follows that any dyadic may be expressed as the sum of three dyads, and so, that either the antecedents or the consequents shall be any desired non-complanar vectors, but only in one way when either the antecedents or the consequents are thus given.

In particular, the dyadic

${\begin{aligned}aii&+bij+cik\\+a'ji&+b'jj+c'jk\\+a'ki&+b''kj+c''kk,\end{aligned}}$

which may for brevity be written

$\scriptstyle {\left\{{\begin{matrix}\ \\\\\ \\\ \\\ \\\ \ \end{matrix}}\right.}$	$a$⁠	$b$⁠	$c$	$\scriptstyle {\left.{\begin{matrix}\ \\\\\ \\\ \\\ \\\ \ \end{matrix}}\right\}\,}$,
	$a'$	$b'$	$c'$
	$a''$	$b''$	$c''$

is equal to

$\alpha i+\beta j+\gamma k,$

where

${\begin{aligned}\alpha &=ai+a'j+a''k,\\\beta &=bi+b'j+b''k,\\\gamma &=c'+c'j+c''k,\end{aligned}}$

and to

$i\lambda +j\mu +k\nu ,$

where

$\alpha i+\beta j+\gamma k,$

where

${\begin{aligned}\lambda &=ai+bj+ck,\\\mu &=a'i+b'j+c'k,\\\nu &=a''+b''j+c''k,\end{aligned}}$

119. By a similar process, the sum of three dyads may be reduced to the sum of two dyads, whenever either the antecedents or the consequents are complanar, and only in such cases. To prove the latter point, let us suppose that in the dyadic

$\alpha \lambda +\beta \mu +\gamma \nu$

neither the antecedents nor the consequents are complanar. The vector

$\{\alpha \lambda +\beta \mu +\gamma \nu \}.\rho$

is a linear function of $\rho$ which will be parallel to $\alpha$ when $\rho$ is perpendicular to $\mu$ and $\nu ,$ which will be parallel to $\beta$ when $\rho$ is perpendicular to $\lambda$ and $\mu ,$ and which will be parallel to $\gamma$ when $\rho$ is perpendicular to $\lambda$ and $\mu .$ Hence, the function may be given any value whatever by giving the proper value to $\rho .$ This would evidently not be the case with the sum of two dyads. Hence, by No. 108, this dyadic cannot be equal to the sum of two dyads.