Vector space under the real numbers problem

Question

If E is a vector space under the real numbers, with addition and multiplication defined as: $$ u\boxplus v = u+(-v); \space a \boxdot u = (-a)u$$ $$ a,b \in \mathbb{R}; \space u,v \in E $$ How can I prove the multiplication distributive axioms: $$a \boxdot (u\boxplus v) = a\boxdot u \boxplus a \boxdot v$$ $$(a+b)\boxdot u = a\boxdot u \boxplus b \boxdot u$$

Answer 1

You must prove the distributive axioms for the operations defined, i. e., you must show that

$$a\boxdot(u\oplus v) =a\boxdot u\boxplus a\boxdot v$$ which is distributivity of scalar multiplication with respect to vector addition and $$(a+ b)\boxdot u =a\boxdot u\boxplus b\boxdot u$$ which is distributivity of scalar multiplication with respect to field addition .

The proof must come from the definition:

$$a\boxdot(u\boxplus v) = a\boxdot(u-v) = -a(u-v) = -au + av =(-au)\boxplus(-av)= (a\boxdot u)\boxplus (a\boxdot v), $$ and $$ (a+ b)\boxdot u = (-a -b)u = -au+(-bu)=-au\boxplus (bu)=(a\boxdot u)\boxplus (-b\boxdot u). $$

So, the second distributivity rule doesn't apply.

Answer 2

Now $$a \boxdot (u \boxplus v)= (-a)(u +(-v))=((-a)u)+(-((-a)v))=\\ =(a\boxdot u) + (- (a \boxdot v))=(a\boxdot u) \boxplus (a \boxdot v)$$

and similar for the other proof.