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I'm working through Axler's Linear Algebra Done Right (third edition). On p30-31 we are told:

P(F) is the set of all polynomials with coefficients in F

And then it is left to the reader to verify that $P(F)$ is a vector space over F. I'm having a bit of trouble showing that $P(F)$ is closed under addition.

This is not a problem if I can assume that $P(F)$ is the set of all polynomials $p(x)$ with coefficients in $F$ and $x \in F$ . In other words I have no problem showing that $p_{1}(x) + p_{2}(x) \in P(F)$ .

But Axler's definition of $P(F)$ states that it is "the set of all polynomials with coefficients in F." Does this include $p_{1}(x)$ and $p_{2}(z)$ for $x,z \in F$ ? If so then I'm having trouble showing that $p_{1}(x)+p_{2}(z) \in P(F)$ for $x,z \in F$ .

I'm still wrapping my head around vector spaces of functions. Is there a convention with respect to the functions' arguments that I'm not aware of? (E.g., when discussing vector spaces of functions is it assumed that the argument of the functions is the same for all functions in that space? I.e. $p_{1}(x), p_{2}(x), p_{3}(x)...$ or can we assume the space includes $p_{1}(x), p_{2}(y), p_{3}(z)...$ ?)

Any guidance here would be appreciated.

NAD
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    Your reference should clarify that they mean polynomials in one variable, $x$ (assuming that is what they mean). But be careful...they aren't asking you to show that sum of values is in the field, they want you to show that the sum of two polynomials is again a polynomial. Thus with $p_1(x)=2x^3-x^2+3x+1$ and $p_2(x)=x^2+5x+2$ we get $(p_1+p_2)(x)=2x^3+8x+3$ confirming the result in this particular case. – lulu Oct 05 '23 at 00:10
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    I would read the definition as meaning the set of polynomials in a single variable. But even if you permit different variables, you're still fine. For example, $x+z$ is a polynomial in the variables $x$ and $z$. – Robert Shore Oct 05 '23 at 00:11
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    Note that you should not view polynomials as functions, either! – Hagen von Eitzen Oct 05 '23 at 15:12

3 Answers3

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You omitted the surrounding context of the statement defining $\mathcal P(\mathbf F)$, which first makes clear the nature of these polynomials as functions of one argument and afterwards the way in which one defines their addition:

Our next example of a vector space involves polynomials. A function $p : \mathbf F \to \mathbf F$ is called a polynomial with coefficients in $\mathbf F$ if there exist $a_0,\ldots,a_m \in \mathbf F$ such that $$ p(z) = a_0 + a_1 z + a_2 z^2 +\ldots+ a_m z^m $$ for all $z\in \mathbf F$. We define $\mathcal P(\mathbf F)$ to be the set of all polynomials with coefficients in $\mathbf F$. Addition on $\mathcal P(\mathbf F)$ is defined as you would expect: if $p, q \in \mathcal P(\mathbf F)$, then $p + q$ is the polynomial defined by $$ (p + q)(z) = p(z) + q(z)$$ for $z ∈ \mathbf F$.

(Linear Algebra Done Right, Second Edition Sheldon Axler, Ch. 1, pg. 10)

The distinction between (1) polynomials as functions of one argument and (2) formal polynomials of a single indeterminate is subtle but important. The author limits consideration of the field $\mathbf F$ to either the real or complex numbers:

Throughout this book, $\mathbf F$ stands for either $\mathbf R$ or $\mathbf C$.

(ibid. pg. 3)

When the field $\mathbf F$ is of characteristic zero, it is infinite and an equivalence can be made between the polynomials functions $\mathcal P(\mathbf F)$ and the formal polynomials with coefficients in $\mathbf F$ of a single unknown (indeterminate). Whether that variable is $x$ or $z$ makes no difference in the case of polynomial functions and only a superficial difference in the case of formal polynomials.

For your purpose, proving closure of polynomials under addition, the definition of adding functions guarantees that we get a sum of two functions as being a function. The difficulty then lies in showing that this function (from $\mathbf F$ to $\mathbf F$) can be represented in the required form. I'd guess the author expects the reader to be satisfied with a demonstration that "collects like terms", i.e. coefficients of equal powers of the argument $z$.

hardmath
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  • Thanks. I'm working from the Third edition which has a slightly different discussion of polynomials in Chapter 2. I don't fully understand the following comment of yours but will certainly investigate as it seems relevant (though perhaps it will become clear as I make more progress in the book): "The distinction between (1) polynomials as functions of one argument and (2) formal polynomials of a single indeterminate is subtle but important." – NAD Oct 05 '23 at 01:42
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    @NAD: I am not happy that the book confuses polynomials with polynomial functions. If $\mathbf F$ were the field $\mathbb F_2$ with two elements, then there would be only four polynomial functions $p\colon \mathbf F\to\mathbf F$ because there are only two choices for $p(0)$ and two for $p(1)$. But the space $\mathbf F[X]$ of polynomials in one variable $X$ is infinite (e.g., there is a distinct $X^n$ for every natural number $n$). Viewed as polynomial functions $\mathbb F_2\to \mathbb F_2$, the polynomials $X$ and $X^2$ are however indistinguishable. – Hagen von Eitzen Oct 05 '23 at 15:40
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    I'm taking a look at the third edition to see how its discussion of polynomials might differ from the second edition. In any case I think a comparison of how the two approaches to defining "polynomials" affects the proofs of closure under addition will illuminate that distinction. – hardmath Oct 06 '23 at 02:16
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A "polynomial with coefficients in $F$" is just the sequence of its coefficients. We usually write that as a formal sum of powers of a "variable" $x$, but there is no element $x$ in $F$. It does not matter what you call the variable.

See What actually is a polynomial?

Ethan Bolker
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  • I am not sure that I understand your comment "there is no element in ." I do see how a polynomial can be represented as a list of its coefficients, but I'm not sure I grasp how this is equivalent to the function itself (particularly in the context of thinking about vector spaces of functions). But I will certainly think about this and do some more reading about polynomials. (I'm still early in the book and this is my first exposure to linear algebra beyond doing arithmetic on vectors and matrices in applied contexts). – NAD Oct 05 '23 at 01:46
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    In this exercise you are not and need not think of the polynomials as functions. They are just expressions you can add and multiply by scalars. The usual rules for addition turn out to be just what you need to check the vector space algorithms. (You can also multiply polynomials, but that's irrelevant when you're thinking just in vector space terms.) Later on in your study you may well see that evaluating polynomials at some particular value is in fact an important thing in vector space terms. Just not yet. – Ethan Bolker Oct 05 '23 at 02:15
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    @EthanBolker: hardmath’s answer points out that in fact the Axler book OP is working from does define polynomials as certain functions, not as formal expressions. In a different context this answer would be the right one, but it doesn’t fit the definition used in that book (which — as you know, but worth noting for OP — is a slightly old-fashioned definition of polynomials, with some disadvantages, but plenty of people still like it, so it’s not a bad one to stick with while you’re first working through this material). – Peter LeFanu Lumsdaine Oct 05 '23 at 16:19
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    @PeterLeFanuLumsdaine I agree. If polynomials are functions of a particular kind then the addition of polynomials is just a special case of the addition of functions. Then in order to show that the polynomials form a subspace of the function space you have to know/assert (or implicitly invoke) the obvious well known algorithm for adding polynomial functions. – Ethan Bolker Oct 06 '23 at 00:18
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Perhaps one way to solve this (i.e. to show that $P(F)$ is closed under addition) is as follows (though I worry that I'm glossing over something more fundamental here since this side-steps the answer from Ethan above in which he points out that "there is no element $x$ in $F$):

Axler writes:

A function $p:F\rightarrow F$ is called a polynomial with coefficients in $F$ if there exist $a_{0},...,a_{m} \in F$ such that $ p(z)=a_{0}+ a_{1}z + a_{2}z^2 + ...+a_{m}z^m $ for all $z \in F$.

So if we have $p_{1}(z), p_{2}(x) \in P(F)$ we'd need to show that $p_{1}(z) + p_{2}(x) \in P(F)$.

So to start: $p_{1}(z)+ p_{2}(x) = p_{2}(x)+a_{0}+ a_{1}z + a_{2}z^2 + ...+a_{m}z^m$.

We can set $b_{0} = p_{2}(x)+a_{0}$. And we know that $b_{0}\in F$ (because $a_{0}\in F$ and $p_{2}(x)\in F$ and $F$ is closed under addition). So then we have $p_{1}(z)+ p_{2}(x) = b_{0}+ b_{1}z + b_{2}z^2 + ...+b_{m}z^m$ which is a polynomial with coefficients in $F$ and therefore it's in $P(F)$ so $P(F)$ is closed under addition.

Is this right? Or should I not be thinking about the arguments to $p$ at all?

NAD
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    Indeed thinking about the argument of $p$ is important, regarding the way Axler presents the polynomials as a "different" example of a vector space (differing from a list of coordinates). – hardmath Oct 05 '23 at 01:40