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 $3$

  • $\begingroup$ Hint: Consider $n=2$, and always send {2} to the origin. What is the resulting dimension of that vector space of functions? It equals the dimension of $V$ where $n=1$. Now note that you can send {2} instead to any point in the 3-dimensional space $\mathbb{R}^3$. $\endgroup$ – Chris Gerig May 8 '18 at 21:55

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.)

  • $\begingroup$ how does the functions look like then? $\endgroup$ – DRPR May 8 '18 at 23:05
  • $\begingroup$ Maybe one function (for $n=3$) sets $f(1) = (0,\pi, 17)$ and $f(2) = (3,-3,\sqrt2)$ and $f(3) = (9000, -1, 1)$. The coordinates for each $f(k)$ really can be arbitrary. $\endgroup$ – Misha Lavrov May 8 '18 at 23:08
  • $\begingroup$ I am finding it a little difficult to accept your explanation, since till now i have only dealt with linear transformations. Here is a different question .Can the elements of V be written in the form of $3 \times 1$ matrix?. Is E a vector space? If so then will its dimension be n? $\endgroup$ – DRPR May 8 '18 at 23:31
  • $\begingroup$ No, they can't, and no, it isn't. (You could write the elements of $V$ in the form of $3 \times n$ matrices, but it would only be a way of bookkeeping; there wouldn't be any meaningful thing to do with matrix multiplication.) $\endgroup$ – Misha Lavrov May 8 '18 at 23:34
  • $\begingroup$ got it .thank you $\endgroup$ – DRPR May 8 '18 at 23:42

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.

  • $\begingroup$ $(a_{11},a_{12},a_{13})$ can be wriiten as a linear combinations of { $e_1,e_2,e_3 $}??Why do we need that extra"n"? $\endgroup$ – DRPR May 8 '18 at 22:43
  • $\begingroup$ You are correct. poor choice of notation on my part. $\endgroup$ – Mohammad Riazi-Kermani May 9 '18 at 1:27

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