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I recently started reading Linear Algebra Done Right by Axler, and I find it great up until Notation 1.23, where the first bullet point states:

If $S$ is a set then $F^S$ denotes the set of functions from $S$ to $F$

I would really like to know what this means; in other words, does this mean that for all $f \in S$, then $f \in F$, but the only difference is that $F$ is a vector field?

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    $\begingroup$ It doesn't mean what you're suggesting. It means that $F^S$ is precisely the set of all functions that have the form $f:S \to F$, where the domain of $f$ is all of $S$ but the range of $f$ is contained within $F$ (but need not be all of $F$). $\endgroup$ Feb 21 at 0:46
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    $\begingroup$ Side advice: the Axler book is titled in an insurrectionist and sensational way. As someone who had to use it in a first course for linear algebra, there is nothing "done right" about his method. Traditional treatments of the subject are better, say by Friedberg. $\endgroup$
    – Favst
    Feb 21 at 0:50
  • $\begingroup$ @Favst, if there are better resources, then which ones would you personally suggest to use? $\endgroup$ Feb 21 at 0:57
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    $\begingroup$ @JonathanDavis speaking only for myself, I prefer the Axler treatment to many others. If you are more interested in matrix theory then Strang or Hoffman/Kunze are almost certainly better, although to be quite honest it is a long time since I have worked with either of them. $\endgroup$ Feb 21 at 1:56
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    $\begingroup$ @Favst You state that you had to use Linear Algebra Done Right for a first course in linear algebra but "there is nothing done right" about the method. The first sentence on the back cover of the book states that this book is intended "for a second course in linear algebra". Thus it is not surprising that you had a bad experience, because in your case you were the wrong audience for the book. $\endgroup$ Feb 21 at 6:28
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The question has already been answered, but here is a little context: the notation $Y^X$ is a little odd to denote the set of functions $X \rightarrow Y$. The origin of this notation lies in combinatorics, and more generally set theory.

Let $[n]$ be the set $\{1,\dots, n\}$. It is a classic motivational exercise in combinatorics to count the size of the set $\{f | f: [n] \rightarrow [m]\}$, and it turns out the answer is $m^n$ many functions (can you prove this?). So when $X$ and $Y$ are finite, we have the wonderful correspondence:

$$\left| Y^X\right| = |Y|^{|X|}$$

Moreover, once you have some cardinal arithmetic under your belt, we may extend this correspondence to infinite sets as well.

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    $\begingroup$ Thank you for letting me know, I was a bit curious about where the notation may have come from! $\endgroup$ Feb 21 at 1:24
  • $\begingroup$ Great answer. For what it's worth, I dislike this notation outside of finite sets and/or a combinatorical or set theoretical context. It's just too many things to potentially misremember or have to think about. $\text{Functions}(X,Y)$ would be better, for heaven's sake. $\endgroup$ Feb 21 at 1:59
  • $\begingroup$ @leslietownes I agree. I always write the domain first, and then have to erase it! For that matter, the closest thing we have to yours is $\text{Hom}(X,Y)$, and even then it's not obvious that we denote the set of morphisms by a thing called "$\text{Hom}$." $\endgroup$
    – William
    Feb 21 at 2:21
  • $\begingroup$ @William absolutely, once people write $\text{Hom}$ it seems like the notation is putting me in the world where morphisms can be almost anything you want them to be. $\endgroup$ Feb 21 at 2:23
  • $\begingroup$ I thought it was a functional analysis notation, generally, though I know little about the subject. Though you didn't say it wasn't. $\endgroup$
    – user403337
    Feb 21 at 2:48
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It means that if $f\in \mathbb{F} ^S$ then $f$ is a function from the set $S$ to a field (think $\mathbb{R}$ or $\mathbb{C}$ for example), $f:S\to\mathbb{F} $. Also, $\mathbb{F} ^S$, with the sum (sum of functions) and product by scalar (scalar times function) as defined in 1.23 is a vector space.

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