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Let $F$ be the set of all complex-valued functions $f:X \to \mathbb{C}$ defined on the set $X =${1, 2, 3} with 3 elements. Show that $F$ is a vector space over the field $\mathbb{R}$ of real numbers. Describe an isomorphism between $F$ and the real vector space $\mathbb{R}^n$, for some integers to be determined. Is this isomorphism unique?

I tried to prove $F$ is a vector space by checking 8 conditions. And then I do not know how to construct a isomorphism between $F$ and the real vector space $\mathbb{R}^n.$

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  • $\begingroup$ Try and solve the question first when $X = \{ 1 \}$ has only one element. Can you show that it is a real vector space? What should be the dimension? $\endgroup$ – levap Apr 17 '17 at 20:38
  • $\begingroup$ You might first want to see things from a broader perspective in order to show there is a vector space structure on $F$. Try to prove: If $S$ is a set and $n \in \mathbb{N}$, the set $$ F_S:=(\mathbb{R}^n)^S \quad\overset{(def.)}{:=} \quad\{f;\;f \;\text{is a map from}\; S \rightarrow \mathbb{R}^n\} $$ inherits a vector space structure from $\mathbb{R}^n$ (e.g., what map would make sense as a sum of two such maps? Ask yourself a similar question for the scalar product.). After that, put $S:=X$ (so $F_S=F_X$) and consider the obvious isomorphy $\mathbb{C}\cong \mathbb{R}^2$ $\endgroup$ – polynomial_donut Apr 17 '17 at 20:56
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Note that defining a function $f:X\to \mathbb C$ is the same as defining the values of $f(1)$, $f(2)$ and $f(3)$, that is, three complex numbers. So the obvious mapping here is

$\phi: F\to \mathbb C^3, \phi(f)=(f(1), f(2), f(3))$

But we're working over the field $\mathbb R$, so we'd like to write $\mathbb C^3$ in the form $\mathbb R^n$ for a certain natural $n$. Now, recall that any complex number $z=x+iy$ can be written as a pair $(x,y)$. So we now can rewrite the mapping above like this:

$\phi: F\to \mathbb R^6, \phi(f)=(Re\ f(1), Im\ f(1), Re\ f(2), Im\ f(2),Re\ f(3), Im\ f(3))$

So, to finish, you just have to show that $\phi$ is an isomorphism.

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