Dropping a rule in vector spaces

[eric_h22]

What happen if i drop the
"V is a vector spaces. If k,l \in \mathbb{F}, u \in V, (k+l)u=ku + lu"
rule in vector spaces?

 

[HallsofIvy]

Well, the first thing that goes is the whole notion of scalar multiplication and representation in terms of bases or as n-tuples of numbers!

If lu+ ku is not equal to (l+ k)v then scalar multiplication no longer has the usual properties. And while we may still be able to write $u= a_1e_1+ a_2e_2$ and $v= b_1e_1+ b_2e_2$ we could no longer say $u+ v= (a_1+ b_1)e_1+ (a_2+ b_2)e_2$.

 

[Erland]

We should also ask ourselves if this axiom is dependent of the other, that is, if it can be proved from the other vector space axioms, and thus being redundant.

The answer is no, it is in fact independent of the other axioms. To see this, we construct a structure which satisfies all vector space axioms except the one in question here.

To do this, we use the real numbers both as our vectors and our field of scalars (with its ordinary field structure).

We define addition of vectors as ordinary addition of reals, but we define multiplication (we denote it with x here) of a vector v by a scalar r by

r x v = v if r>0, r x v = -v if r<0, and r x v = 0 if r=0.

It is then easy show that all the vector space axioms are satisfied for this structure, except the axiom in question here, which is false here, since (1+1) x 1 = 2 x 1 = 1, while 1 x 1 + 1 x 1 = 1 + 1 = 2.

 

[Erland]

A more general problem is to decide which vector space axioms are independent and which can be proved from the other axioms. We found here that the distributivity rule discussed here is independent. In another tread we found that the existence of additive inverses is also independent, although it can be replaced with the rule 0v=v, for all vectors v.

The following I know:Dependent (can be proved from the other axioms)

- Commutativity of vector additionIndependent (cannot be proved from the other axioms):

- Existence of additive inverses

- Vector distrubutes over scalars (the axiom OP asked about)

- 1v=v for all vectorsWhat about the other axioms?

 

[blue_raver22]

If you drop the rule stating that the sum of two vectors multiplied by a scalar is equal to the sum of each individual vector multiplied by that scalar, you would no longer have a vector space. This rule, also known as the distributive property, is essential in defining vector spaces and is one of the fundamental properties that sets vector spaces apart from other mathematical structures. Without this rule, the operations of addition and scalar multiplication would not be well-defined, and the axioms of vector spaces would not be satisfied. Therefore, dropping this rule would result in an invalid mathematical structure that cannot be considered a vector space.