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Let $V$ be an arbitrary vector space over $K$. I'm asked to prove that $$a(v_1 + v_2 + ... + v_n) = av_1 + av_2 + ... + av_n$$ $$(a_1 + a_2 + .... + a_n)v = a_1v + a_2v + ... + a_nv$$ for all $a, a_i \in K, \quad v, v_i \in V$.

I'm supposed to do it with induction, but I don't get the question because isn't all this implied with the distributivity axiom of vector spaces, $$c(v+u) = cv + cu; (c_1 + c_2)v = c_1v + c_2v,$$ where $c, c_1, c_2 \in K, \quad v, u \in V.$?

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  • $\begingroup$ That proves the claim for $n = 2$, which does not prove the case for $n > 2$, despite seeming obvious. A bit of extra work is needed. $\endgroup$
    – qualcuno
    Commented Oct 6, 2018 at 22:32

2 Answers 2

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Well, the original axiom is about only two vectors

You are asked to use induction to generalize to n vectors .

If it seems trivial to you good for you but you still have to do the task of writing a proof.

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Look at the case $n=1$, then this implies that:

$$a(v_{1})=av_{1}$$

which is trivially true. Now assume $\exists k\in N$ for the equation to hold and then follow through the algebra.

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