Finding Dimension of a vector space $V$ Let $E = \{1, 2,.....,n\}$, where $n$ is an odd positive integer. Let $V$ be
the vector space of all functions from $E$ to $\mathbb{R^3}$, where the vector space operations are given by 
$$(f+g)(k)=f(k)+g(k);   f, g \in V;  k \in E$$
$$ (\lambda f)(k)= \lambda f(k);f\in V;  k \in E$$
Find the dimension of $V$
First of all, I know that the vector space $V$ is finite since the number of functions from a finite set $E$ to set  $\mathbb{R^3}$ is finite. And under the operation given above, I've shown that $V$ is a vector space. But I've no idea in finding out the dimension of $V$. Should I construct one linear transformation from V to some other vector space, and then use the rank-nullity theorem to find out the dimension of $V$. Suggest any help.
 A: We could do this for just $\mathbb{R}^3$ but we might as well generalize it to $\mathbb{R}^m$.
Let $\{e_1,e_2,\dots,e_m\}$ be a basis of $\mathbb{R}^m$. Define $\varphi_{i,j} : E \to \mathbb{R}^m$ with $1\leq i\leq n$ and $1\leq j\leq m$ by $\varphi_{i,j}(k) = \delta_{ik}e_j$ (where $\delta_{ij}$ is the Kronecker delta).
The set of all such $\varphi_{i,j}$ (which we will call $B$) spans $(\mathbb{R}^m)^E$ because if $$\varphi(k) = \sum_{j=1}^m\lambda_{k,j}e_j$$ for each $k\in E$ and some scalars $\lambda$, then 
$$
\varphi=\sum_{i=1}^n\sum_{j=1}^m\lambda_{i,j}\varphi_{i,j}
$$
To see this, note that 
$$
\begin{align*}
\sum_{i=1}^n\sum_{j=1}^m\lambda_{i,j}\varphi_{i,j}(k) &= \sum_{i=1}^n\sum_{j=1}^m\lambda_{i,j}\delta_{ik}e_j \\
&= \sum_{j=1}^m\lambda_{k,j}e_j \\
&= \varphi(k)
\end{align*}
$$
Now, to prove that $B$ is linearly independent, note that every $\varphi_{i,j}$ sends each $i \in E$ to $e_j$. If some linear combination of $B$ is the zero function, that would imply that some linear combination of $e_j$ (with the same corresponding scalars!) must be zero, meaning the scalars must be zero. Hence $B$ is a basis of $(\mathbb{R}^m)^E$, and $\dim (\mathbb{R}^m)^E = mn$.
