# Axiom 10 of Vector Spaces

• I
So I understand how to prove most of the axioms of a vector space except for axiom 10, I just do not understand how any set could fail the Scalar Identity axiom; Could anybody clarify how exactly a set could fail this as from what I know that anything times one results in itself

1u = u
1(x,y,z)=(x,y,z)
1(1,2,3) = (1,2,3)
1 (1,0,...,1) = (1,0,...,1)

I don't see how you could ever end up in a situation where you could end up with

1VV

Mark44
Mentor
So I understand how to prove most of the axioms of a vector space except for axiom 10, I just do not understand how any set could fail the Scalar Identity axiom; Could anybody clarify how exactly a set could fail this as from what I know that anything times one results in itself

1u = u
1(x,y,z)=(x,y,z)
1(1,2,3) = (1,2,3)
1 (1,0,...,1) = (1,0,...,1)

I don't see how you could ever end up in a situation where you could end up with

1VV
It can happen if you have a vector space with an unusual kind of scalar multiplication. Keep in mind that a vector space consists of a set of vectors over some field (often, the field of real numbers ##\mathbb{R}## or the field of complex numbers ##\mathbb{C}##), together with operations for vector addition and for multiplication by a scalar.

If scalar multiplication is defined like this
##k \cdot <x, y> = <kx, 0>##
then there is no scalar k for which ##k \cdot <x, y> = <x, y>##.