If $V$ is a vector space, then $V^{n}$ too? Let $n \in \mathbb{N}$. Let $V$ be a vector space. Is $V^n$ a vector space too?
Thanks in advance.
 A: Yes. You should be able to write down the addition and scalar multiplication rules. The dimension will be $n$ times the dimension of $V$.
For example, If $V = \mathbb{R}^2$ and $n=3$ then $V^n$ is the set of ordered triples of ordered pairs, which is essentially just $\mathbb{R}^6$.
More abstractly, if $V$ is any vector space then $V^n$ is the set of ordered $n$-tuples of elements of $V$. You add those $n$-tuples by adding each coordinate, just the way you build $\mathbb{R}^n$ from the one dimensional vector space $\mathbb{R}$ .
A: I don't mean to be pedantic at all but I think the correct way to answer this is: If $V$ is a vector space over $\mathbb{K}$ then there is a natural way to make $V^n = V \times \cdots \times V$ into a vector space in which we define: 
$$(v_1,...,v_n) + (w_1,...,w_n) := (v_1+w_1,...,v_n+w_n)$$
$$ \hspace{-1.1in}\lambda (v_1,...,v_n) := (\lambda v_1,...,\lambda v_n), \forall \lambda \in \mathbb{K}$$
and with this, one checks that $V^n$ becomes a vector space over $\mathbb{K}$ as well. In other words, alone $V^n$ is just a set, but we can endow $V^n$ with a vector space structure.  
