I am attempting to work my way through the 3rd edition of "Linear Algebra Done Right", but there's a paragraph on page 14 that I don't understand. I have struggled with it myself for a few hours and have come to the conclusion that I need some help. But first some background.

Axler lets $\textbf{F}$ stand for either the set of real or complex numbers. (He mentions fields but doesn't use them directly.) He also uses the term "list" instead of "tuple".

The problematic paragraph is preceded by this:

  • If $S$ is a set, then $\textbf{F}^S$ denotes the set of functions from $S$ to $\textbf{F}$.
  • For $f, g \in \textbf{F}^S$, the sum $f + g \in \textbf{F}^S$ is the function defined by $$(f + g)(x) = f(x) + g(x)$$ for all $x \in S$.
  • For $\lambda \in \textbf{F}$ and $f \in \textbf{F}^S$, the product $\lambda f \in \textbf{F}^S$ is the function defined by $$(\lambda f)(x) = \lambda f(x)$$ for all $x \in S$.

As an example of the notation above, if $S$ is the interval [0,1] and $\textbf{F} = \textbf{R}$, then $\textbf{R}^{[0,1]}$ is the set of real-valued functions on the interval [0,1].

So far, so good. This is where I get lost:

Our previous examples of vector spaces, $\textbf{F}^n$ and $\textbf{F}^\infty$, are special cases of the vector space $\textbf{F}^S$ because a list of length $n$ of numbers in $\textbf{F}$ can be thought of as a function from {1, 2, ..., $n$} to $\textbf{F}$ and a sequence of numbers in $\textbf{F}$ can be thought of as a function from the set of positive integers to $\textbf{F}$. In other words, we can think of $\textbf{F}^n$ as $\textbf{F}^{\{1,2,...,n\}}$ and we can think of $\textbf{F}^\infty$ as $\textbf{F}^{\{1,2,...\}}$.

The general idea seems straightforward to me. Let $S$ be the set of all tuples in $\textbf{F}^n$ or $\textbf{F}^\infty$, and now $\textbf{F}^S$ is a vector space. Is it really that simple? But what is he saying?

[My background: I am a computer programmer who took a "normal" linear algebra class in college, who has a new-found love for higher mathematics, but not enough time and money to go back to school.]

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    $\begingroup$ Do you understand what it means for a list or a sequence of numbers to be represented by a function from $\mathbb{N}$ to your field? $\endgroup$ – qbert Mar 12 '16 at 22:28
  • $\begingroup$ Everything you write seems ok except that last paragraph (Let $S$ be the set of all tuples...) I'm not sure what you are trying to say. $\{1,\ldots,n\}$ is simply an $n$-element set, not the space of tuples. $F^n=$ "all functions from $\{1,...,n\}$ to $F$". "n-tuples" are elements of $F^{\{1,...,n\}}$, not $\{1,\ldots, n\}$. But of course, you can take $S$ to be all of $F^n$ and then $F^S$ is another (infinite dimensional) vector space. $\endgroup$ – Peter Franek Mar 12 '16 at 22:29
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    $\begingroup$ S isn't the set of n-tuples. S is the set of just the numbers 1 through n. Let f be a function from S to R. Example f(1)=-27, f(2)=$\pi$, f(3)=21/38, f(4) = $\sqrt{e}$ etc. This one function is the same thing as the one n-tuple (-27,$\pi$,21/38, $\sqrt{e}$....). The function is the n-tuple. The set of all n-tuples is the vector space $\mathbb R^n$. The set of all real-valued functions with domain {1,2,3,4...n} the map into the real numbers is $\mathbb R^{\{1,...n\}}$. The point is $\mathbb R^n = \mathbb R^{\{1..n\}}$. Functions are n-tuples; n-tuples are funcctions. $\endgroup$ – fleablood Mar 12 '16 at 22:51

It's actually simpler but much more subtle than that.

Let $\mathbb R^{\{1...n\}}$ is the set of all functions from $\{1,2...,n\} \rightarrow \mathbb R$.

For example if $f(x) = x^2 - 3$ then $f\in \mathbb R^{\{1...n\}}$.

But $f$ can be thought of as an n-tuple. If, for instance, $n= 3$.

$f = (-2,1,6) \in \mathbb R^3$

If $f(x) = e^x$ then $f \in \mathbb R^{\{1,2,3\}} = (e, e^2, e^3) \in \mathbb R^3$.

In other words: Every 3-tuple $(x_1, x_2, x_3)$ in the vector space $\mathbb R^3$ can be thought as a function $f:\{1,2,3\} \rightarrow \mathbb R$ where $f(1) = x_1; f(2) = x_2; f(3) = x_3)$. The set of functions and the set of n-tuples are the same thing.

  • $\begingroup$ It was your comment above that made it all click for me. (Thank you very much for your help!) $\endgroup$ – sam.bishop Mar 12 '16 at 23:30

I think I should just assure you that sequences $x_n$ do make up a vector space, they can be added termwise and multiplied by a constant.

A less obvious example: Let us take a sequence of rational numbers $x_n$ such that $$ x_{n+2} = x_{n+1} + x_n. $$ This is a vector space over $\mathbb Q,$ and is of dimension exactly two. You can add them and get another, you can multiply by a rational number and get another such sequence. Not sure if you have read about bases yet, anyway, here are two basis vectors for the sequence: $$ y_1 = 1, \; \; y_2 = 1, \mbox{then} \; \; y_{n+2} = y_{n+1} + y_n, $$ $$ z_1 = 1, \; \; z_2 = 3, \mbox{then} \; \; z_{n+2} = z_{n+1} + z_n. $$ The first is the Fibonacci numbers, the second the Lucas numbers. Any sequence following the recursion is a linear combination of these.


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