Polynomial equal to polynomial of lower degree

Question

I am studying Linear Algebra Done Right, chapter 2 problem 6 states:

Prove that the real vector space consisting of all continuous real valued functions on the interval $[0,1]$ is infinite dimensional.

My solution:

Consider the sequence of functions $x, x^2, x^3, \dots$ This is a linearly independent infinite sequence of functions so clearly this space cannot have a finite basis. However this prove relies on the fact that no $x^n$ is a linear combination of the previous terms. In other words, is it possible for a polynomial of degree $n$ to be equal to a polynomial of degree less than $n$. I believe this is not possible, but does anyone know how to prove this? More specifically, could the following equation ever be true for all $x$?

$x^n = \sum\limits_{k=1}^{n-1} a_kx^k$ where each $a_k \in \mathbb R$

José Carlos Santos · Answer 1 · 2017-08-06T14:53:07.280

12

Then the polynomial $\displaystyle x^n-\sum_{k=0}^{n-1}a_kx^k$ would have infinitely many roots, but it can have $n$, at most.

Another way of dealing with this problem is based upon defining polynomials (in one variable $x$) as expressions of the type $a_0+a_1x+a_2x^2+\cdots+a_nx^n$, where $n\in\{0,1,2,\dots\}$ and each $a_n$ is real. Under this definition, the polynomial $a_0+a_1x+a_2x^2+\cdots+a_nx^n$ is equal to the polynomial $b_0+b_1x+a_2x^2+\cdots+b_nx^n$ if and only if the coefficients are equal, that is, if and only if $a_0=b_0$, $a_1=b_1$, and so on. Under this definition, the problem discussed here is trivial.

What did I prove above then? Well, for each $P(x)\in\mathbb{R}[x]$, there is a corresponding polynomial function from $\mathbb R$ into $\mathbb R$. What I proved above is that this correspondence is one-to-one — when we are dealing with $\mathbb R$. It is still one-to-one if we are dealing with any field with charactristic $0$, such as $\mathbb Q$ or $\mathbb C$. But this is not true in general. For instance, if our field is $\mathbb{F}_2$, then $x$ and $x^2$ are distinct polynomials. But they correspond to the same polynomial function.

edited Aug 06 '17 at 14:53

answered Aug 06 '17 at 14:06

José Carlos Santos

427,504

2

"Equations" can have infinitely many solutions, but nonzero polynomials cannot. – Bill Dubuque Aug 06 '17 at 14:11
Thank you I just need to find a proof of that fact. If you have any good references that would be great. – Joe Aug 06 '17 at 14:12
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@Joe A reference? Why? Isn't my proof clear? – José Carlos Santos Aug 06 '17 at 14:14
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@Joe See this answer for a proof of the root bound for polynomials. – Bill Dubuque Aug 06 '17 at 14:16
A proof of the fact that a polynomial of degree n has at most n solutions. – Joe Aug 06 '17 at 14:16
@BillDubuque I have edited my answer. – José Carlos Santos Aug 06 '17 at 14:17
1

@Joe You can find several proofs of the Fundamental theorem of algebra. – Sahiba Arora Aug 06 '17 at 14:18
@Joe See here. – José Carlos Santos Aug 06 '17 at 14:19
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@BillDubuque I think you mean "a proof for the root bound for nonzero polynomials" :P – Zubin Mukerjee Aug 06 '17 at 14:19
1

@Joe If you know about unique factorization of polynomials over fields then a slick proof is that if a nonzero polynomial has infinitely many roots $r_i$ then it has infinitely many nonassociate prime factors $,x - r_i,,$ contra unique factorization. – Bill Dubuque Aug 06 '17 at 14:29
1

In Linear Algebra Done Right, Sheldon Axler states on page 24 that the fact that a polynomial of degree m has at most m roots will be proved in chapter 4. I think maybe I should just wait until chapter 4. Are there other ways to prove the original question without this fact about polynomials? – Joe Aug 06 '17 at 14:37
@Joe Please read my expanded answer. – José Carlos Santos Aug 06 '17 at 14:53
@JoséCarlosSantos But it seems like you are going around the problem by changing the definition of a polynomial? – Joe Aug 06 '17 at 14:58
3

@ZubinMukerjee Jokes aside, I really do mean what I wrote. We should expect statements of results to be precise, but not names of results (else we'd be stuck with monstrous names like "the root bound for nonzero formal univariate polynomials over a commutative integral domain"). – Bill Dubuque Aug 06 '17 at 14:59
@BillDubuque That makes sense! I hadn't thought about that distinction before. – Zubin Mukerjee Aug 06 '17 at 15:03
1

@Joe I just wanted to make you aware of the distinction between polynomial and polynomial function. Perhaps that Axler assumed that his readers knew that a poynomial with degree $n$ has, at most, $n$ roots. Or perhaps that he had in mind a solution without poynomials. For instance, he might have been thinking about functions of the type $f_n(x)=(x-n)(1-x+n)$, which are zero outside the interval $(n,n+1)$ and non-zero inside. Therefore, they are linearly independent. – José Carlos Santos Aug 06 '17 at 15:07

score 6 · Answer 2 · edited Jul 23 '19 at 17:52

6

By definition, $x^n + \sum\limits_{k=0}^{n-1}a_k x^k$ is a nonzero polynomial of degree $n$. A nonzero polynomial of degree $n$ can have at most $n$ zeros in any field, and $[0,1]$ has more than $n$ elements.

edited Jul 23 '19 at 17:52

darij grinberg

17,777

answered Aug 06 '17 at 14:11

Jonas Meyer

53,602

score 5 · Answer 3 · answered Aug 06 '17 at 14:56

5

Two identical functions have identical derivatives. Differentiate your equality $n$ times and you will see the two sides cannot be identical.

answered Aug 06 '17 at 14:56

Kamil Maciorowski

2,818

Good point, but I have not studied derivatives on a rigorous level yet... so I can't really use that as part of my proof. – Joe Aug 06 '17 at 15:01

score 0 · Answer 4 · answered Aug 06 '17 at 20:42

Suppose $x^n = \sum\limits_{k=1}^{n-1} a_kx^k$. Consider what happens when $x$ is much larger than $|a_1|+\ldots+|a_{n-1}|$. By the triangle inequality, you will get a contradiction. Essentially, you are showing that $x^n$ grows faster than any degree $n-1$ polynomial...

Polynomial equal to polynomial of lower degree

4 Answers4