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I found this abstract algebra question in a previous test paper:

Suppose $\Bbb{F}$ is a field and $\mathrm{X}$ is a non-empty set. Then $\text{Maps}(\mathrm{X},\Bbb{F})$ is a vector space over $\Bbb{F}$.

  • If $|\mathrm{X}|=3$ then find a canonical basis of $\text{Maps}(\mathrm{X},\Bbb{F})$.

  • Can you recognize $\text{Maps}(\mathrm{X},\Bbb{F})$ where $\mathrm{X}=\{1,2,3\}\times \{1,2\}$ & $\Bbb{F}=\Bbb{R}$ ?

I'm not sure what $\text{Maps}$ mean here? Is it a standard term in abstract algebra? Also, what exactly is meant by "canonical" basis? Moreoever, what is meant by $|\mathrm{X}|$? Does it refer to the number of elements in the set $\mathrm{X} $?

For the second case is $\mathrm{X}$ just the Cartesian Product of $\{1,2,3\}$ and $\{1,2\}$, or no?

Any suggestion regarding how to approach this question is appreciated.

3 Answers3

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The notation $\operatorname{Maps}(X,\mathbb{F})$ is not “standard”, but it can be easily understood: the set consists of all maps $X\to\mathbb{F}$.

You can endow it with pointwise addition, $f+g\colon x\mapsto f(x)+g(x)$, and pointwise scalar multiplication, $af\colon x\mapsto af(x)$.

A “canonical basis” (the name is not well chosen, in my opinion) consists of the maps that have value $1$ at a single point of $X$ and $0$ on all others.

A map $\{1,2,3\}\times\{1,2\}\to\mathbb{F}$ associates to each pair $(i,j)$ (with $1\le i\le 3$ and $1\le j\le 2$) an element of $\mathbb{F}$: does this ring a bell?

Matrices! This is an abstract definition for them.

egreg
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  • You mean like: A matrix has an element $a_{ij} \in \Bbb{F}$ for any given pair $(i,j)$ ? So, is it like $\operatorname{Maps}(X,\mathbb{F})$ is the set of all $3\times2$ matrices? I guess the trivial basis, in that case, would be formed by ${{1,0},{0,0},{0,0}}$, ${{0,1},{0,0},{0,0}}$ and so on (6-element basis). –  Apr 19 '18 at 22:15
  • And similarly, for the $|\mathrm{X}|=3$ case, the canonical basis would be formed of $3$ elements? –  Apr 19 '18 at 22:18
  • @JJD Yes, that's true. – egreg Apr 19 '18 at 22:25
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Yes, $|X|$ means the number of elements of $X$. And "maps" means "functions". The basis is "canonical" because it is so obvious and natural.

Also $\times$ in this context denotes the Cartesian product.

Martin Argerami
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  • Thanks, but I'm not sure I can understand which "so obvious and natural basis" you are referring to. Could you please elaborate? BTW in the second case $X$ is the cartesian product ${(1,1),(1,2),(2,1),(2,2),...}$ right? –  Apr 19 '18 at 22:05
  • Yes, it's the cartesian product. As for the basis, did you find any basis of $\operatorname{Maps}(X,\mathbb F)$? – Martin Argerami Apr 19 '18 at 22:07
  • I guess it is ${{1,0},{0,0},{0,0}}$, ${{0,1},{0,0},{0,0}}$ and so on (i.e. a 6 element basis), for the second part of the question ? –  Apr 19 '18 at 22:20
  • Yes, if $X$ is finite, you get a canonical basis by considering the characteristic functions $1_{{x}}$ for each $x\in X$, – Martin Argerami Apr 19 '18 at 22:32
  • Can you now agree that it's "obvious and natural"? ;) – Martin Argerami Apr 19 '18 at 22:34
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Since each $f \in \text{Maps}$ is determined by the values $f(a_1),f(a_2),f(a_3)$ in the field, the natural basis for $\text{Maps}$

$$x_i(a_j)= \delta_{ij}$$

As a quick check, for any $f \in \text{Maps}$ with $\alpha_j=f(a_j)$ look at

$$q(t)=\sum\limits_j \alpha_jx_j(t)$$ Then note that, $$q(a_j)=\alpha_j=f$$

It remains to show that the $x_j$'s are linearly independent which then defines a basis for $\text{Maps}$.

Dinesh
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