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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: $$a \boxdot (u\boxplus v) = a\boxdot u \boxplus a \boxdot v$$ $$(a+b)\boxdot u = a\boxdot u \boxplus b \boxdot u$$

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  • $\begingroup$ Don't you want to prove $a \boxdot (u \boxplus v)=a \boxdot u \boxplus a \boxdot v$ and $(a \boxplus b) \boxdot v= a \boxdot v \boxplus b \boxdot v$? $\endgroup$
    – user289143
    Apr 26, 2020 at 19:42
  • $\begingroup$ Corrected the post, thank you $\endgroup$
    – user707245
    Apr 26, 2020 at 19:47

2 Answers 2

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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.

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  • $\begingroup$ Thank you, this makes it clear for me! $\endgroup$
    – user707245
    Apr 26, 2020 at 21:46
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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.

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  • $\begingroup$ Thanks for the explanation! $\endgroup$
    – user707245
    Apr 26, 2020 at 21:46

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