Question

Vector space question:
Can the scalar multiplication identity be defined as something other than 1 to meet the axiom?
For example, if V is the set of vectors x=(x_1,x_2) where vector addition is defined as normal and scalar multiplication is defined as ax=(2ax_1,2ax_2), would (1/2)x=(x_1,x_2)=x satisfy the axiom?

Answer 1

if u given any set of vectors defined, ur scalar multiplication must not change ur value of x

Answer 2

the scalar identity has to do with the field of scalars not with the vectors.

if u have a field F in which "1/2" is the multiplicative neutral element, then for an abelian group V to be a vector space over F then (1/2)v=v for every vector v.

Answer 3

The question doesn't specify a field so I will assume that the scalars are all real numbers. And since the multiplicative identity of real numbers is 1, the scalar identity should be 1 and this vector space fails this axiom?

Answer 4

yes.

but since the axioma does not hold. what u wrote namely
\[ \large \alpha(x_1;x_2)=(2\alpha x_1;2\alpha x_2) \]
is not a scalar multiplication hence u don't have a vector space.