kx1 y1 z1 kz1 ky1 kx1 Do these operations form a vector spa

k.(x₁, y₁, z₁) = (kz₁, ky₁, kx₁)

Do these operations form a vector space?

Find all the axioms that are satisfied an prove them. Provide counterexample for all axioms that fail. (Hint: 6 Axioms FAIL)

Solution

(x,y,z) + (u,v,w) = (z+w, y+v,x+u)

(1) Axioms which are satisfied under the operations

(a) closed under addition

(b) closed under scalar multiplication

(c) associativity of the addition operation

(d) distributivity of scalar multiplication over addition

(ii) Axioms not satisfied under these operations

(a) existence of identity: If (x,y,z) + (a,b,c) = (z+c,y+b,x+a) = (x,y,z) for all x,y,z then

                                                                       z+c =x and x+a =z

                                                                But a and c will depend on x and z,

so there is no identity element for the addition operation.

Hence this operation fails to satisfy the group axiom on addition

Consequently, the existence of inverse will also not be satisfied.

1(x,y,z) is not equal to (x,y,z)