6—Vector Spaces 146B~
A~
A~+B~
A~
2 A~
(A~+B~) +C~ A~+ (B~+C~)
The associative law, axiom 3, is also illustrated in the picture. The zero vector, axiom 4, appears in this
picture as just a point, the origin.
The definition of multiplication by a scalar is that the length of the arrow is changed (or even reversed) by
the factor given by the scalar. Axioms 7 and 8 are then simply the statement that the graphical interpretation of
multiplication of numbers involves adding and multiplying their lengths.
A~′
A~
α(A~+B~)
Axioms 5 and 9 appear in this picture.
Finally, axiom 10 is true because you leave the vector alone when you multiply it by one.
This process looks almosttooeasy. Some of the axioms even look as though they are trivial and unnecessary.
The last one for example: why do you have toassumethat multiplication by one leaves the vector alone? For an
answer, I can show you an example of something that satisfies all of axioms one through nine butnotthe tenth.
These processes, addition of vectors and multiplication by scalars, are functions. I could write “f(~v 1 ,~v 2 )” instead
of “~v 1 +~v 2 ” and write “g(α,~v)” instead of “α~v”. The standard notation is just that — a common way to write
a vector-valued function of two variables. I can define any function that I want and then see if it satisfies the
required properties.
On the set of arrows that I just worked through, redefine multiplication by a scalar (the functiong) to be the
zero vector for all scalars and vectors. That is,α~v=O~for allαand~v. Look back and you see that this definition
satisfies all the assumptions 1–9 but not 10. This observation proves that the tenth axiom is independent of the
others. If you could derive the tenth axiom from the first nine, then this example couldn’t exist. This construction