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 $3$
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.)