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 $x$? (Except $n=0$ where $f(x)$ is $0$ in all cases)
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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. |
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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). |
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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 $f$, $f=(x-a)q$ wehere $q$ 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. $\Box$ |
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