Let $E = \{1,2,\ldots,n \}$, where n is a positive integer. Let $E = \{1,2,\ldots,n \}$, where $n$ is a positive integer. 
Let $V$ be the vector space of all functions from $E$ to $\mathbb{R^3}$
My question is how can the dimension of vector space $V$ be $3n$ ?
The matrix of transformations should look somewhat like 
$$\begin{bmatrix}   a \\b \\c \\\end{bmatrix} \begin{bmatrix} k \\ \end{bmatrix} =\begin{bmatrix}   x_1 \\x_2 \\x_3 \\\end{bmatrix} \text{where k }\in E$$
from what i see dimension of vector space should be 3 , but it isnt . Can anyone explain to me where am i going wrong?
$ a \begin{bmatrix}   1 \\0 \\0 \\\end{bmatrix} + b \begin{bmatrix}   0 \\1 \\0 \\\end{bmatrix} +c \begin{bmatrix}   0 \\0 \\1 \\\end{bmatrix}= \begin{bmatrix} a \\b \\c \\\end{bmatrix}$ 
Thus the basis of $V$ is $\{e_1,e_2,e_3\}$ $\Rightarrow $ dimension is $3$
 A: You are thinking of functions $E \to \mathbb R^3$ that are linear. (That is, linear if extended to functions $\mathbb R \to \mathbb R^3$.) It is those functions that can be written as $$\begin{bmatrix}   a \\b \\c \\\end{bmatrix} \begin{bmatrix} k \\ \end{bmatrix} =\begin{bmatrix}   x_1 \\x_2 \\x_3 \\\end{bmatrix} \text{where }k\in E.$$
In such cases, once you choose $f(1)$ (for which you have three choices, hence the space is $3$-dimensional), then $f(2), f(3), \dots, f(n)$ are determined for you by linearity.
But if you consider arbitrary functions, then you can independently choose $3n$ real numbers: the $3$ coordinates of $f(1$), then the three coordinates of $f(2)$, and so on up through the three coordinates of $f(n)$. So the vector space is $3n$-dimensional.
(Similarly, the vector space of linear functions $\mathbb R \to \mathbb R^3$ is $3$-dimensional, but the vector space of all functions $\mathbb R \to \mathbb R^3$ has uncountably many dimensions. Just a very small part of the possibilities in the second case - functions whose each coordinate is a polynomial of arbitrary degree - is already infinite-dimensional.)
A: A function from $\{1,2,...,n\}$ to $\mathbb {R}^3$ assigns $(a_{11},a_{12}, a_{13})$ to $1$ and  $ (a_{21},a_{22}, a_{23})$ to $2$, ...,  $ (a_{n1},a_{n2}, a_{n3})$  to $n$.
Thus there are $3n$ degrees of freedom involved, which makes it a $3n$ dimensional vector space.   
