27
$\begingroup$

How can I prove, that a polynomial function $f(x) = \sum_{0\le k \le n}a_k x^k\qquad n\in\mathbb N,\ a_k\in\mathbb C$ is zero for at most $n$ different values of $x$, unless all $a_0,a_1,\ldots,a_n$ are zero?

  • 0
    @Adrián: Not really sure, as perhaps one could give complex analysis proofs of this. The only reason I talked about fields was to try and point out the potential inaccuracy of the statement "It's a part of the fundamental theorem".2011-03-08

8 Answers 8

-1

Let $ f $ be a polynomial, such that $ f(r) = 0 $. Define $ f'(x) = \frac{f(x)}{x - r} $. Repeat this process, and consider the degree, $ \deg(f^{(n)}) $.

  • 0
    It is so on purpose. I would simply want to point OP in a direction.2016-06-15
53

You don't need the fundamental theorem of algebra or the Vandermonde determinant, only the factor theorem.

Proposition: A polynomial of degree at most n with more than n roots vanishes identically.

Proof: By induction. The base case is $n=0$, which is obvious. Now take a polynomial f of degree at most n, and let $x_1,\ldots,x_{n+1}$ be distinct roots of f. By the factor theorem, we can write $f(x) = (x-x_{n+1})g(x)$ where g plainly has degree at most $n-1$. Now substitute $x = x_i$ for $i=1,\ldots,n$. For all these values of x the left hand side vanishes and the factor $(x_i-x_{n+1})$ is nonzero. Hence all these $x_i$ must be roots of g and by induction g is identically zero. QED

This same proof works over any field (or even integral domain).

  • 0
    for polynomials $f,g\in K[x]$ there are $q,r\in K[x]$ such that $f=qg+r$ with \text{deg}(r)<\text{deg}(g) or $r=0$ (polynomial division). if $f(\alpha)=0$, take $g(x)=x-\alpha$ to get $0=f(\alpha)=r$. Hence $r=0$ and $f(x)=q(x)(x-\alpha)$.2011-03-08
28

The following literature may be of use here:

Theorem. A polynomial $\text{f}$ of degree $\text{n}$ over a field $\text{F}$ has at most $\text{n}$ roots in $\text{F}$.*

Proof. The results is obviously true for polynomials of degree $0$ and degree $1$. We assume it to be true for polynomials of degree $n-1$. If $a$ is a root of $f$, $f=(x-a)q$ where $q$ has degree $n-1$. Since $f(b)=0$ if and only if $a=b$ or $q(b)=0$, it follows by our inductive assumption that $f$ has at most $n$ roots. $\Box$

  • 0
    The theorem is **not** true for polynomials of degree 0! The zero polynomial has degree 0, but it has infinitely many roots if $F$ is infinite.2018-10-18
27

Using Linear Algebra,

If the $n+1$ distinct roots are $\alpha_i$, then we have that $x = [a_0, a_1, \dots, a_n]^{T}$ is a solution of $Ax = 0$ where $A$ is the Vandermonde matrix using the $\alpha_i$.

Since the Vandermonde matrix is invertible for distinct $\alpha_i$, it follows that $x = [0, 0, \dots, 0]$.

Thus if $a_j \neq 0$ for some $j$, then your polynomial can have at most $n$ different roots.

Note: This is basically saying that given a field $K$, any polynomial of degree $n$ in $K[x]$ has at most $n$ distinct roots.

Fundamental Theorem of Algebra is an assertion of the fact that $\mathbb{C}$ is algebraically closed, and the $K$ above need not be algebraically closed.

  • 0
    @PranavBisht: No. There is an explicit formula for the determinant of the Vandermonde matrix. It would actually be an interesting proof that uses the theorem in question to prove that the Vandermonde matrix is invertible...2017-04-04
3

Just a clarification here. The Fundamental Theorem of Algebra says that a polynomial of degree n will have exactly n roots (counting multiplicity). This is not the same as saying it has at most n roots. To get from "at most" to "exactly" you need a way to show that a polynomial of degree n has at least one root. Then you can proceed by induction.

There are lots of different kinds of proofs that a polynomial must have at least one root. None of them are totally trivial.

1

Suppose that the polynomial is $$f=\sum_{i=0}^na_iX^i \in \mathbb{C}\left[X\right] .$$ Assume that the polynomial function defined by the above polynomial has $n+1$ distinct roots, i.e., $f(x)=0$ for $n+1$ distinct values of x $\in \Bbb C$.
I shall use the theorem:$\prod_{i=1}^k\left(X-x_i\right)q=f \Longleftrightarrow x_1,...,x_k \text{ are distinct roots of }f\text{, where }q\in\Bbb C[X] $ Let $n+1$ distinct roots of $f$ are $x_1,...,x_{n+1}$. Hence, by above theorem, we can say that $ \prod_{i=1}^{n+1}\left(X-x_i\right)q=f$. Since the coefficient lie in the field of complex numbers, hence left hand side has degree $n+1$, while $f$ is of degree n. This is impossible, therefore $f$ can have at most n distinct roots.

1

Let $\alpha_1,\ldots,\alpha_{n+1}$ be distinct roots for $f(x)$ with ${\sf deg}(f)\leq n$. W.l.o.g, we can assume the leading coefficient of $f(x)$ is always $1$. This can be done by enumerating all the cases:

If ${\sf deg}(f)=n$, then $\alpha_1,\ldots,\alpha_n$ are roots implies $f(x)=(x-\alpha_1)\cdots(x-\alpha_n)$. But $f(\alpha_{n+1})\neq0$, so this does not work.

If ${\sf deg}(f)=n-1$, then $\alpha_1,\ldots,\alpha_{n-1}$ are roots implies $f(x)=(x-\alpha_1)\cdots(x-\alpha_{n-1})$. But $f(\alpha_n)\neq0$, so this does not work either.

$\vdots$

If ${\sf deg}(f)=1$, then $\alpha_1$ is a root implies $f(x)=x-\alpha_1$. But $f(\alpha_2)\neq0$, so this does not work either.

Hence this only case that would work is ${\sf deg}(f)=0$, and so $f(x)=0$.

  • 0
    @FUZxxl This is a consequence of factor theorem: $\alpha$ is a root of $f(x)$ iff $f(x)=(x-\alpha)g(x)$ where ${\sf deg}(g)={\sf deg}(f)-1$. And note that $\alpha_2$ is a root of $(x-\alpha_1)g(x)$ implies $\alpha_2$ is a root of $g(x)$...2016-11-29