Page:Linear Algebra (1882) Tevfik.djvu/21

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7. The operation of tracing the line $AB$ from the point $A$, and the line $BD$ from the point $B$, and giving to them their respective directions, will be represented by the expression $AB+BD$; and to show that by this operation $AD$ is found, the expression $AB+BD=AD$ will be employed.

8. If in departing from the point A the lines $AB,BD,DH,HN,NO$, are successively traced in their respective directions, the line joining $A$ to $O$, or $AO$, will be represented by

$AO=AB+BD+DH+NO$.

It is readily seen that after having traced AB, if in place of tracing the other lines in the order given, we trace successively a line parallel and equal in length to each one of these lines, in their respective directions, in whatever order, we shall still find the same line A O. It is needless to say that this manner of representation of straight lines is general.

9. It is now apparent what in Linear Algebra is meant by $AB+BD=AD$. If the lines $AB,BD$ are found equal in length, it is evident the length of $AD$ will diminish with the angle $ABD$; and finally $AD$ will become zero whenever this angle does; in this case the point $D$ coincides with $A$, and the line $BD$ with $BA$; for this reason

$AB+BD=0$ or $AB+BA=0$.

Thus in the expressions

$AB+BA$ or $BA+AB$

$AB$ and $BA$ neutralize each other; therefore when a line measured in one direction is represented by a positive symbol, the same line measured in the opposite direction may be represented by the same symbol taken negatively, that is

$AB=-BA$ or $BA=-AB$,

hence if the line AB is represented by $\rho$, the line $BA$ will be $-\rho$.

10. If $AB,DE$ are on the same right line, and in the same direction, we admit, as in Numerical Algebra, that $AB$ is to $DE$ as ${\underline {AB}}$ to ${\underline {DE}}$, that is

$AB={\frac {\underline {AB}}{\underline {DE}}}DE$.

Now if ${\underline {AB}}={\underline {DE}}$, then $AB=DE$ and consequently

$AB+ED=0$.

11. If $AB,DE$ are parallel in the same direction, and ${\underline {AB}}={\underline {DE}}$, we must admit

$AB=DE$.

For if we take $AN,DM$ on the same right line $AD$, and ${\underline {AN}}={\underline {DM}}$, we admit $AN=DM$ (art. 10),

but $DE$ compared to $DM$ is situated exactly as $AB$ compared to $AN$, and this similarity of position is so complete that if we know $AB$ from its relation to $AN$ it will be exactly