Prove that the following is vector space

Question

Prove that for any $n\ge 1$, $(K^n/K,+,\cdot)$ is a vector space, where $(K^n,+)$ is an abelian group and $(K,+,\cdot)$ is a commutative field.

I've been having trouble proving this lately. Any help is welcome!

Answer 1

Abstract: yes, it is.

In order to prove it, we need to prove the following:

1. Distributivity of $v \in K^n $ over $t_1, t_2 \in K $. This is often called "Pseudo-Distributivity".
2. Distributivity of $t \in K $ over $v_1, v_2 \in K^n $.
3. Associativity of $v \in K^n $ over $t_1, t_2 \in K $.
4. Normalization. $1\cdot v = v$ for all $v \in K^n$.

I assume $K$ is a numerical set. I believe the definition of summation and product by scalar in $K^n$ is already given, so I'll assume those too.

1. Let $v \in K^n$, $t_1, t_2, t_3 \in K$. Let $t_3 = t_1, t_2$. Then: $$ v(t_1 + t_2) = t_3v $$ $$ t_1v + t_2v = t_3v $$ $$ \implies v(t_1 + t_2) = t_1v + t_2 v $$
2. Let $v_1, v_2, v_3 \in K^n$, $t \in K$. Let $v_3 = v_1 + v_2$. Then: $$ t(v_1 + v_2) = tv_3 $$ $$ tv_1 + tv_2 = tv_3 $$ $$ \implies t(v_1 + v_2) = tv_1 + tv_2 $$
3. Let $v \in K^n$, $t, s, p \in K$. Let $p = ts$. Then: $$ (ts)v = (p)v = pv $$ $$ t(sv) = pv $$ $$ \implies (ts)v = t(pv) $$
4. Let $v \in K^n$. Then: $1\cdot v = v$.

These properties follow from how operations on your vector set are defined. To put it briefly, summation should be the usual memberwise summation of vectors, while multiplication by scalar should be the usual memberwise product. You will find out that under these circumstances, your vector space $(\mathbb{K}^n, \mathbb{K}, +, \cdot )$ is the typical coordinate vector space over $\mathbb{K}$. Of course, this works for every $n \in \mathbb{N}$. An interesting insight about this is that a commutative field can always be considered a vector space over itself. You could also prove this by induction, given the above insight.